mathematic center d96 munirka village … 12th/class...mathematic center d96 munirka village new...

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1. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩∩∩∩ F) = 0.2, find

P (E|F) and P(F|E)

• Solution: Given P(E) = 0.6, P(F)= 0.3, P(E ∩ F) = 0.2

P(E / F) = = =

P(F / E) = = =

2. Compute P(A|B), if P(B) = 0.5 and P (A ∩∩∩∩ B) = 0.32

• Solution: P(B) = 0.5 P(A ∩ B) = 0.32

P(A/B) = = = 0.64

3. If P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4, find

(i) P(A ∩∩∩∩ B) (ii) P(A|B)

• Solution: P(A) = 0.8 P(B) = 0.5 P(B/A) = 0.4

(i) P(A ∩ B) = P(B/A) × P(A) = 0.4 × 0.8

= 0.32

(ii) P(A/B) = = = 0.64

4. Evaluate P(AUB), if 2P(A) = P(B) = and P(A|B) =

• Solution:

2P(A) = P(B) =

MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315

CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA, DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi EMAIL:1 [email protected] web site www.mathematic.in

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P(A/B) =

P(A) = , P(B) =

P(A ∩ B) = P(A/B) – P(B)

= × =

P(A ∪ B) = P(A) + P(B) – 2 P(A ∩ B)

=

=

= =

5. If P(A) = , P(B) = P(A ∪∪∪∪ B) = find (i) P(A ∩∩∩∩ B) (ii) P(A|B) (iii) P(B|A)

• Solution:

P(A) = , P(B) = P(A ∪ B) =

(i) P(A ∩ B) = P(A)+ P(B) – P(A ∪ B)

=

=

(ii) P(A/B) = = =

=



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(iii) P(B/A) = = = =

6. A coin is tossed three times, where

(i) E : head on third toss , F : heads on first two tosses

(ii) E : at least two heads , F : at most two heads

(iii) E : at most two tails , F : at least one tail

• Solution: A coin tossed 3 times is same as three coins tossed.

S = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH}

n(S) = 8

E: heads on third toss.

E = {TTH, HTH, THH, HHH}

P(E) = = =

F: heads on first two tosses.

F = {HHT, HHH}

P(F) = = =

E ∩ F = {HHH}

n(E ∩ F) = 1

P(E/F) = = =

ii) E = At least two heads

E = {HHT, HTH, THH, HHH}



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n(E) = =

F = {TTT, HTT, THT, TTH, HHT, HTH, THH}

N(F) = 7

E ∩ F = {HHT, HTH, THH}

N(E ∩ F) = 3

P(E ∩ F) = =

P(E / F) = = ÷ =

(iii) E: At most two tails

E: {HTT, THT, TTH, HHT, HTH, THH, HHH}

n(E) = 7

P(E) = =

F: At least one tail.

F = {TTT, HTT, THT, TTH, HHT, HTH, THH}

n(F) = 7

n(F) = =

E ∩ F = {HTT, THT, TTH, HHT, HTH, THH}

n(E ∩ F) = 6

P(E ∩ F) =

P(E / F) = = =



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7. Determine (E/F): Two coins are tossed once, when

(i) E : tail appears on one coin F : one coin shows head

(ii) E : no tail appears, F : no head appears

• Solution: (i) E : tail appears on one coin F : one coin shows head

S = {HH, TH, HT, TT}

n(S) = (4)

E : tail appears on one coin.

E: {TH, HT}

n(E) = 2

P(E) = = =

F: one coin shows head.

F = {TH, HT}

n(F) = 2

p(F) = = =

E ∩ F = {TH, HT}

n(E ∩ F) = 2

P(E ∩ F) = =

Thus P(E/F) = = = 1

(ii) E : no tail appears, F : no head appears

E = [H, H}

n(E) = 1



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p(E) = =

F(TT)

n(f) = 1

P(f) = =

E ∩ F = φ

N(E ∩ F) = 0

N(E ∩ F) = 0

P(E/f) = = 0

8. A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears

respectively on first two tosses.

• Solution: A dice has 6 faces (ie) 1, 2, 3, 4, 5, 6

n(s) = 6 × 6 × 6 = 216

E = (1, 2, 3, 4, 5, 6) × (1, 2, 3, 4, 5, 6) × 4

F = (6) × (5) × (1, 2, 3, 4, 5, 6)

n(f) = 1 × 1 × 6 = 6

P(F) = =

E ∩ F = {6, 5, 4}

P(E ∩ F) = =

P(E / f ) = = =



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9. Mother, father and son line up at random for a family picture E : son on one end, F : father in middle.

• Solution: S = {MFS, MSF, SFM, SMF, FMS, FSM}

n(S) = 6

E = son on one end

= [MFS, SFM, SMF, FMS}

n(E) = 4

F = Father in middle

F = {MFS, SFM}

n(F) = 2

p(F) =

E ∩ F = {MFS, SFM} = 2

n(E ∩ F) = 2

p(E/F) = = = 1

10. A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

• Solution: (a) n(s) = 6 × 6 = 36

Let A represents obtaining a sum greater than 9 and B represents black die resulted in a s.

A {46, 64, 55, 36, 63, 45, 54, 65, 56, 66}

n(A) = 10



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P(A) =

B = {51, 52, 53, 54, 55, 56}

n(B) = 6

p(B) =

A ∩ B = {55, 56}

n(A ∩ B) = 2

p(A ∩ B) =

P(A / B) = = = =

b) Let A denote the sum is 8.

∴ A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

B = Red die results in a number less than 4.

Either first or second die is red.

B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5)

(1, 6), (2, 1), (2, 2), (2, 3), (2, 4)

(2, 5), (2, 6), (3, 1), (3, 2), (3, 3)

(3, 4), (3, 5), (3, 6)}

A ∩ B = {(2, 6), (3, 5)]

P(A ∩ B) = =

P(B) = =

Hence P(A / B) = = = =



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11. A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}

Find

(i) P(E|F) and P(F|E) (ii) P(E|G) and P(G|E)

(ii) P[(E U F) | G] and P[(E ∩∩∩∩ F) | G]

• Solution: S = {1, 2, 3, 4, 5, 6}

n(S) = 6

E = {1, 3, 5}

F = {2, 3}

G = {2, 3, 4, 5}

n(E) = 3, n(F) = 2, n(4) = 4 (i) To find P(E/F) and P(F/E)

P(E) = =

P(F) = =

E ∩ F = [3]

n( E ∩ F) = 1

p(E ∩ F) = =

p(E / F) = = =

and P(F/E) = = =

(ii) To find P(E/G) and

P(G/E)



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P(E) =

P(G) = =

E ∩ G = {3. 5}

n(E ∩ G) = 2

P(E ∩ G) = =

P(E/G) = = = =

P(G/E) = = × =

(iii) To find P[(E ∪ F) / G] and P((E ∩ F)/G]

P(G) =

E ∪ F = {1, 2, 3, 5}

G = {2, 3, 4, 5}

(E ∪ F) ∩ G = {2, 3, 5}

n[(E ∪ F) ∩ G] = 3

P[(E ∪ F) ∩ G] =

P[(E ∪ F)/G] = =

=

Again E ∩ F = {3}



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(E ∩ F) ∩ G = {3}

n[(E ∩ F) ∩ G] = 1

P[E ∩ F / G] = = =

12. Assume that each born child is equally likely to be a boy or a girl. If a family has two

children, what is the conditional probability that both are girls given that

(i) the youngest is a girl, (ii) at least one is a girl?

• Solution: Let the first and the second girls are denoted by G1 and G2 and Boys be B1 and B2.

Sample space

S = {(G1, G2), (G1, B2), (G2, B1), (B1, B2)

Let

A = Both the children are girls

= {G1, G2}

B = Youngest child in a girls

= {G1, G2, B1G2]

c = at least one girl in a girl

= {G1B2, G1G2, B1G1]

A ∩ B = {G1G2}

A ∩ C = {G1, G2}

P(A ∩ B) =

P(A ∩ C) =

P(B) = , p(C) =



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(i) p(A/B) =

= =

(ii) p(A/C) = = ÷

=

13. An instructor has a question bank consisting of 300 easy True / False questions, 200

difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

• Solution: Test bank has

Number of easy True / False questions = 300

Number of difficult True / False questions = 200

Number of easy multiple choice questions = 500

Number of easy difficult multiple choice questions = 400

Total Number of all such questions n(s) = 1400

Let ‘E’ represents an easy questions and F represents a multiple choice question.

n(E) = 300 + 500 = 800

&#n(F) = 500 + 400 = 900

To find P(E/F)

P(F) = =

&#n(E ∩ F) = 500

n(E ∩ F) = =



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P(E / F) =

= =

14. Given that the two numbers appearing on throwing two dice are different. Find

the probability of the event ‘the sum of numbers on the dice is 4’.

• Solution: The sample spare on throwing two dice

S =

n(s) = 36

Let A represent the event "the sum of numbers on the dice is 4" and B represents the event" the two numbers appearing on throwing two dice are different" A = {13, 22, 31}

n(A) = 3

p(A) = =

for ‘B’ n(B) = 30

p(B) = =

n(A ∩ B) = 2



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P(A ∩ B) =

Hence the required probability = P(A/B) = =

= =

15. Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die

again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

• Solution: n(S) for throwing two dice = 36

Let ‘A’ represent at least one die showing 3.

A = {(1,3),(2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3)}

n(A) = 11

P(A) = =

Let ‘B’ represent a number other than 3, then the coin is tossed.

S = {1H, 2H, 4H, 1H, 1T, 2T, 4T, 5T}

= 8

A ∩ B = {13, 31}

n(A ∩ B) = 2

n(A ∩ B) = =

P(A / B) = = = =

16. If P(A) = p(B) = , find P (A ∩∩∩∩ B) if A and B are independent events.



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• Solution:

Given P(A) = p(B) = .

As A and B are independent events,

∴ P(A ∩ B) = P(A) P(B)

= =

17. Two cards are drawn at random and without replacement from a pack of 52 playing

cards. Find the probability that both the cards are black.

• Solution: S = {52 cards]

n(s) = 52

Two cards are drawn without replacement

A = {26, black cards}

n(A) = 26

n(A) =

p(A and B) = P(A) P(B)

= × = ×

=

18. A box of oranges is inspected by examining three randomly selected oranges drawn

without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

• Solution: S = {12 good oranges, 3 bad oranges}

n(s) = 15

P(a box is approved)



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= =

=

19. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the

coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

• Solution: S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

n(s) = 12

A = {H1, H2, H3, H4, H5, H6}

n(A) = 6

p(A) = = =

B = {H3}

n(B) = 1

p(B) = =

A ∩ B = {H3}

n(A ∩ B) = 1

p(A ∩ B) = =

Again P(A). p(B) = =

∴ P(A ∩ B) = P(A). P(B)

Hence A and B are independent events.

20. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the

number is even,’ and B be the event, ‘the number is red’. Are A and B independent?



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• Solution:

S =

n(S) = 6

A = {2, 4, 6}

n(A) = 3

p(A) = = =

B =

n(B) = 3

p(B) = = =

A ∩ B = [2]

n(A ∩ B) = 1

p(A ∩ B) = =

Now, P(A). P(B) = =

Thus, P(A ∩ B) ≠ P(A) . P(B)

21. Let E and F be events with P(E) = P(F) = and P(E ∩∩∩∩ F) = . Are E and F independent?

• Solution:

P(E) = P(F) =

P(E ∩ F) =

Now P(E).R(F) = =



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Now P(E ∩ F) ≠ P(E). P(F)

22. Given that the events A and B are such that P(A) = , P(A ∩∩∩∩ B) = and

P(B) = p. Find p if they are (i) mutually exclusive (ii) independent.

• Solution:

P(A) = , P(A ∩ B) =

p(B) = p

(i) A and B are mutually exclusive events.

∴ P(A ∪ B) = P(A) + P(B)

=

P = -

= =

(ii) A and B are independent events.

∴ P(A ∩ B) = P(A) + P(B) – P(A) . P(B)

=

- = P

P =

P =

23. Let A and B be independent events with P (A) = 0.3 and P(B) = 0.4. Find

(i) P(A ∩∩∩∩ B) (ii) P(A U B)



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(iii) P (A|B) (iv) P (B|A)

• Solution: P(A) = 0.3 p(B) = 0.4

A and B are independent events.

(i) ∴ P(A ∩ B) = P(A). P(B)

= 0.3 × 0.4 = 0.12

(ii) P(A ∪ B) = P(A) + P(B) – P(A) . P(B)

= 0.3 + 0.4 – 0.3 × 0.4

= 0.7 – 0.12 = 0.58

(iii) P(A/B) = =

= 0.3

(iv) P(B/A) = =

= 0.4

24. If A and B are two events such that P (A) = P (B) = and P (A ∩∩∩∩ B) = find P (not A and not B).

• Solution:

P(A) = P(B) =

P(A ∩ B) =

P(not A) = 1 – P(A) – 1 - =

P(not B) = 1 – P(B) – 1 - =

Now P(A).P(B) = =



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∴ p(A ∩ B) = P(A). P(B)

Thus A and B are independent events.

So not A and not B are independent events.

Here P(not A and not B)

= P(not A). P(not B)

= × =

25. Events A and B are such that P (A) = , P(B) = and P(not A or not B) = .

P( = 1 – P(A ∩∩∩∩ B)

= 1 – p(A ∩∩∩∩ B)

• Solution:

P(A ∩ B) = 1 - =

P(A) =

P(B) =

P(A). P(B) =

=

P(A ∩ B) =

P(A ∩ B) ≠ P(A) × P(B)

26. Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6.

Find (i) P(A and B) (ii) P(A and not B) (iii) P(A or B) (iv) P(neither A nor B)



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• Solution: P(A) = 0.3 p(B) = 0.6 and A and B are independent events.

(i) P(A and B) = P(A) . P(B)

= 0.3 × 0.6 = 0.18

P (A and not B) = P(A ∩ )

= P(A) – P(A ∩ B)

= 0.3 – 0.18 = 0.12

(iii) P(A or B) = P(A) + P(B) + P(A and B)

= 0.3 + 0.6 – 0.18 = 0.72

(iv) P(neither A nor B)

P(not (A ∩ B)] = 1 – P(A ∪ B)

= 1 – 0.72 = 0.28

27. A die is tossed thrice. Find the probability of getting an odd number at least once.

• Solution: S = {1, 2, 3, 4, 5, 6]

n(S) = 6

Let A represents an odd number

A = {1, 3, 5}

n(A) = 3

P(A) = = =

P( ) = 1 – P(A) = 1- =

n = 3

P(atleast one success)



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= 1 – P(x = 0) = 1 -

= 1 - =

28. Two balls are drawn at random with replacement from a box containing 10 black and

8 red balls. Find the probability that

(i) both balls are red.

(ii) first ball is black and second is red.

(iii) one of them is black and other is red.

• Solution: S = [10 Black balls, 8 red balls]

n(S) = 18

Let drawing of a red ball be a success.

∴ A = {8 red balls]

n(A) = 8

P(A) = =

P( ) = 1 - =

(i) P(Both are red balls)

= P(A) . P(A)

= × =

(ii) P(First is black ball and second in red)

∴ P( . P(A)

= =



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(iii) P(one of them is black and other is red)

= P( . P(A) + P(A) . P(

= =

29. Probability of solving specific problem independently by A and B are and respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

• Solution:

P(A) = P(B) =

P( = 1 - =

P( = 1 - =

(i) P(the problem is solved)

- p(the problem is not solved)

= 1 – P(

= 1 – P(

= 1 - × = 1 - =

(ii) P(exacting one of them solves the problem)

= P(A). P( + P( . P(

=

=



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30. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?

(i) E : ‘the card drawn is a spade’

F : ‘the card drawn is an ace’

(ii) E : ‘the card drawn is black’

F : ‘the card drawn is a king’

(iii) E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

• Solution: S = {All the 52 cards}

n(S) = 52

(i) E = {13 spades}

n(E) = 13

n(E) = = =

F = [4 aces}

n(F) = 4

P(F) = = =

E ∩ F = [an ace of spade]

n(E ∩ F) = 1

P(E ∩ F) = =

Also P(E) P(F) = =

P(E ∩ F) = P(E) . P(F)

Here E and F are independent events.

(ii) E = {26 black cards}



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n(E) = 26

P(E) = = =

F = {4 kings} n(F) = 4

P(F) = = =

E ∩ F = {2 Black kings)

n(E ∩ F) = 2

P(E ∩ F) = = =

Now P(E) . P(F) = =

P(E ∩ F) = P(E) . P(F)

Hence E and F are independent events.

(iii) E = {4 kings, 4 queens}

n(E) = 8

P(E) = = =

F = {4 queens, 4 jacks}

n(F) = 8

P(F) = = =

E ∩ F = {4 queens}

n(E ∩ F) = 4

p(E ∩ F) = 4

P(E ∩ F) = = =



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Now P(E), P(F) = × =

Thus P(E ∩ F) ≠ P(E).P(F)

Here E and F are independent events.

31. In a hostel, 60% of the students read Hindi news paper, 40% read English news

paper and 20% read both Hindi and English news papers. A student is selected at random.

(a) Find the probability that she reads neither Hindi nor English news papers.

(b) If she reads Hindi news paper, find the probability that she reads English news paper.

(c) If she reads English news paper, find the probability that she reads Hindi news paper.

Choose the correct answer in Exercises 17 and 18.

• Solution: Let A represents the students reading Hindi news paper are B represents the students reading English news paper.

P(A) = , P(B) =

P(A and B) =

a) P (neither Hindi nor English news paper is read)

= 1 – p(A or B)

= 1 – [P(A) + P(B) – P(A and B)]

= 1 -

= 1 -

(b) P(A ∩ B) =

P(B/A) = =



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= = =

(c) P(A/B) = =

= =

32. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted

and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

• Solution: S1 = {5 red balls, 5 black balls}

n(S1) = 10

Let us draw a red ball first.

A1 = {5 red balls} n(A1) = 5

P(A1) = = =

After adding 2 balls of the same colour, (ie)

S2 = {7 red balls, 5 black balls]

n(S2) = 12

A2 = {7 red balls]

n(A2) = 7

P(A2) = =

P/a red balls is drawn =

=



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Case(ii) when a black ball is drawn

A2 = {5 black balls}

n(A2) = 5

n(A2) = = =

After adding 2 balls of the same colour, (ie) black ball.

S2 = [7 black ball, 5 Red balls]

n(S2) = 12

A2 = {5 red balls}

⇒ n(A2) =

P(a red ball is drawn)

= =

Required probability in both the cases = = =

33. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls.

One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

• Solution: Let A be the event that ball drawn is red and let E1 and E2 be the events that the ball, drawn is from the first bag and the second bag respectively.

P(E1) = P(E2) =

P(A/E1) = Probability of drawing a red ball from bag I

= =

P(A/E2) = Probability of drawing a red ball from bag II



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= =

Therefore by Bayer’s theorem.

P(E1/ A) = Probability that the red ball drawn in from bag I

=

=

= = =

= =

34. Of the students in a college, it is known that 60% reside in hostel and 40% are day

scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

• Solution: Let E1, E2 and A represents the following

E1 = students residing in the hostel

E2 = day scholars (not residing in the hostel)

and A = students who attain grade A.

Now P(E1) =

P(E2) =

P(A/E1) = , P(A/E2) =



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By Bayer’s theorem

P(E1/A) =

=

= = =

35. In answering a question on a multiple choice test, a student either knows the answer

or guesses. Let be the probability that he knows the answer and be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with

probability .What is the probability that the student knows the answer given that he answered it correctly?

• Solution: Let E1, E2 and E3 and A be the events defined as follows:

E1 = the examine knows the answer

E2 = the examine guesses the answer

and A = The examine answers correctly.

Now we have,

P(E1) = P(E2) =

Since E1 and E2 are mutually exclusive and exhaustive. If E2 has already occurred then the examinee guesses, therefore, the probability that the answers correctly given that he has made a guess in

(ie) P(A/E2) =

and P(A/E1) = Probability that the answers correctly given that the knew the answer = 1.

By Baye’s rule.

Required probability.



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= P(E1/ A)

=

= = =

36. A laboratory blood test is 99% effective in detecting a certain disease when it is in

fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?

• Solution: E1 = The person selected is suffering from certain disease.

E2 = The person related is not suffering from certain disease.

A = The doctor diagnoses correctly.

Now P(E1) = 0.1% = = 0.001

P(E2) = 1 - =

P(A/E1) = 99% =

and P(A/E2) = 0.005%

By Baye’s theorem.

P(E1/ A) =

=

=



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37. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

• Solution: Let E1, E2 and A denoted the following:

E1 = a two headed coin

E2 = biased coin

E3 = an unbiased coin

A = A head in shown

P(E1) = P(E2) =

P(E3) =

P(A / E1) = 1 P(A/E2) = =

P(A / E3) =

P(E1 / A) =

=

=

=



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38. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

• Solution: Let E1, E2, E3 and A be the events defined as follows:

E1 = Person chosen is a scooter driver.

E2 = Person chosen is a car driver.

E3 = person chosen is a truck driver.

A = Person meets with an accident.

Since there are 12000 persons therefore

P(E1) = =

P(E2) = =

P(E3) = =

It is given that P(A/E1)

∴ Probability that a person meets with an accidents what is the probability that he was a scooter driver.

By Baye’s theorem

P(E1/A) =

=

=

39. A factory has two machines A and B. Past record shows that machine A produced

60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items



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are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

• Solution: Given

P(A) = ; P(B) =

Let D denote a defective item

∴ P(D/A) =

P(D/B) =

By Baye’s theorem.

P(B/D) =

=

=

40. Two groups are competing for the position on the Board of directors of a corporation.

The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

• Solution: P(G1) = 0.6, P(G2) = 0.4 what P represents the laundry of a new product P(p/G1) = 0.7 p(P/G2) = 0.3 Using Baye’s therem,

P(G2/P) =

= = = =



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41. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and

notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

• Solution: Let E1, E2 and A be the events as mentioned below:

E1 = 5 or 6 appears on a die.

E2 = 1, 2, 3 or 4 appears on a die.

A = A head appears on the coin.

P(E1) = P(E2) =

P(A/E1) = Probability of getting a head on tossing a coin 3 times when E1 already occurs.

= P(HTT) or P(THT) or P(TTH)

=

=

P(A/E2) = Probability of getting a head on tossing a coin. Once, when E2 occurs already.

=

P(E2/A) =

= =

=

42. A manufacturer has three machine operators A, B and C. The first operator A

produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for



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30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

• Solution: E1 the item manufactured by Operator A

E2 the item manufactured by Operator B

E3 the item manufactured by Operator C

and A = the item is defective.

P(E1) = , P(E2) = , P(E3) =

P(A/E1) = probability that the item drawn is manufactured by operator A.

=

P(A/E2) =

P(A/E3) =

Required probability = P(E1/A)

=

=

=

43. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two

cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

• Solution: E1 – the missing card is a diamond

E2 - the missing card is a spade



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E3 - the missing card is a club

E4 - the missing card is a heart

A = Drawing two cards of heart from the remaining cards.

P(E1) =

P(E2) =

P(E3) =

P(E4) =

P(A/E1) = Probability of drawing 2 heart cards given that one diamond card is missing.

P(A/E1) =

P(A/E2) =

P(A/E3) =

P(A/E4) =

∴ P(E1/A) =

=

=



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44. State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

(i)

X 0 1 2

P(X) 0.4 0.4 0.2

(ii)

X 0 1 2 3 4

P(X) 0.4 0.4 0.2 -0.1 0.3

(iii)

Y -1 0 1

P(Y) 0.6 0.1 0.2

(iv)

Z 3 2 1 0 -1

P(X) 0.4 0.4 0.2 0.05 0.05

• Solution: (i) P(O) + P(1) + p(2)

= 0.4 +0.4 + 0.2 = 1

It is a probability distribution

(ii) p(3) = -0.1 which is not possible. Hence it is not a probability distributor.

(iii) p(-1) + p(O) + P(1)

= 0.6 0.1 + 0.2 = 0.9 ≠ 1

Hence not probability distribution

(iv) P(3) +p(2) + p(1) + p(0) + p(-1)

= 0.3 + 0.2 + 0.4 + 0.1 + 0.05

= 1.05 ≠ 1

Hence it is not a probability distribution.



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45. An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?

• Solution: Two balls may be selected as RR, RB, BR, BB where R represents red and b represents black ball.

Variable X, has the 0, 1, 2 (ie) there may be no black ball, may be one black ball (or) both the balls are black.

46. Let X represent the difference between the number of heads and the number of tails

obtained when a coin is tossed 6 times. What are possible values of X?

• Solution: For one coin S = {H, T} (ie) n(S) = 2

Let A represent head A = {H} n(A) = 1

P(A) = P =

n = 6, r = 0, 1, 2, 3, 4, 5, 6

P(x = 0) = [P ]6 = =

P(x = 1) = 6P [P ]5 = 6 × =

P(x = 2) = 15[P ]2 [P ]4 = 15 × =

P(x = 3) = 20[P ]3 [P ]3 = 20 × =

P(x = 4) = 15[P ]4 [P ]3 = 20 × =

P(x = 4) = 15[P ]4 [P ]2 = 15 × =

P(x = 5) = 6[P ]5 [P = 6 × =



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P(x = 6) = [P ]6 = =

47. Find the probability distribution of

(i) number of heads in two tosses of a coin.

(ii) number of tails in the simultaneous tosses of three coins.

(iii) number of heads in four tosses of a coin.

• Solution:

(i) S = {T, H} n(S) = 2

Let A represents the head A= {H}, n(A) = 1

P(A) = =

P( ) = 1- =

Number of tosses

n = 2, r = 0, 1, 2

P(X = ) = P( ) (P( ) = =

P(X = 1 ) = 2 P(A) (P( ) = 2 × =

P(X = 2 ) = P(A) × P(A) = =

Probability distribution

Xi 0 1 2

P(Xi)

(ii) Three coins tossed once is the same one coin tossed 3 times.

S {T, H} n(s) = 2



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Let A represent a tail. A = {T}

n(A) = 1 n(A) =

P( = 1 - =

P(X = 0) = P( .( .(

= × × =

P(X = 1) = 3P( .( .(

= 3 × × × =

P(X = 2) = 3P( .( .(

= 3 × × × =

P(X = 3) = 3P( .P(A)(A) =


Xi 0 1 2 3

Pi

(iii) S = {H, T} n(s) = 2

A represents head A = {H} n(A) = 1

P(A) = =

P( ) = 1 - =

N = 4, r = 0, 1, 2, 3, 4



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N(x = 0) = P( ).P( ) P( )P( )

=

P(x = 1) = 4(P( ) P( ). P( ) - P( )

=

P(x = 2) = 6x P(A) × P( ) × P( ) × P( )

=

P(X = 3) = 4 P(A) P(A) P(A) (P( ))

=

P(X = 4) = P(A) P(A) P(A) (P(A)

=

Yi 0 1 2 3 4

Pi

48. Find the probability distribution of the number of successes in two tosses of a die,

where a success is defined as

(i) number greater than 4

(ii) six appears on at least one die

• Solution: S = {1, 2, 3, 4, 5, 6}

n(s) = 6

(i) Let A be the set of favorable events.

A = {5, 6} n(A) = 2



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n(A) = = =

P( ) = 1 - P(A) = 1 - =

n = 2 r = 0, 1, 2

P(X = 0) = P( ) . P( ) = ×

=

P(X = 1) = 2P(A) P( )

= 2 × × =

P(X = 2) = P(A) P(B) =

Xi 0 1 2

P(d)

(ii) S = {1, 2, 3, 4, 5, 6]

n(S) = 6

Let A represent that 6 appears on one die

A = [6]

n(A) = 1

P = , q =

n = 2 r = 0, 1, 2

P(x = 0) = =



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P(x = 1) = 2 × × =

P(x = 2) = =

P(atleast one six) = +

=

Xi 0 1

P(xi)

49. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at

random with replacement. Find the probability distribution of the number of defective bulbs.

• Solution: n(S) = 30

A = {6 defective bulbs}

n(A) = 6

P = = =

q = 1 – P(A) = 1 - =

n = 4 (4 bulbs are drawn with replacement]

r = 0, 1, 2, 3, 4

P(x = 0) = (q)4 = =

P(x = 1) = 4 c1 p. q3 = 4 × ×

=



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P(x = 2) = 4 c2 p2 q2 = 6 ×

=

p(x = 3)

= 4C3 p3q =

=

P(x = 4) = 4C4 P4 = =


Xi 0 1 2 3 4

P(xi)

50. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is

tossed twice, find the probability distribution of number of tails.

• Solution: Let P represents the appearance of tail.

∴ q represents the appearance of head.

Now q = 3p

P + q = 1

4p = 1

P = 1/4 , q = 3/4

n = 2 (number of tosses)

r = 0, 1, 2

P(x= 0) = 2C0 q2 = =



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P(X = 1) = 2C1 qp = 2 × × =

P(X = 2) = 2C2 P2 = =

Xi 0 1 2

Pi

51. A random variable X has the following probability distribution:

X 0 1 2 3 4 5 6 7

P(X) 0 K 2k 2k 3K k2 2k2 7K2 + k

Determine

(i) k (ii) P(X < 3)

(iii) P(X > 6) (iv) P(0 < X < 3)

• Solution: (i) we know that the sum of all the probabilities of distribution is 1.

P(X = 0) + p(x = 1) + …. P(x = 7) = 1

0 + k + 2k+2k + 3k + k2 + 2k2 + 742 + k = 0

10k2 + 9k – 1 = 0

(10k – 1) (K + 1) = 0

k =

(ii) P(x < 3) = P(O) + P(1) + P(2)

= 0 + + +

(iii) P(X > 6)

= P(7) = 7k2 +k

= 7 +



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= = =

iv) P(O <(X < 3))

p(X = 1) + P(X = 2)

= K + 2k = 3k

= 3 =

52. The random variable X has a probability distribution P(X) of the following form,

where k is some number:

k, if x = 0

2k, if x = 1

3k, if x = 2

0, otherwise

(a) Determine the value of k.

(b) Find P (X < 2), P (X ≤≤≤≤ 2), P(X ≥≥≥≥ 2).

• Solution: The p.d.f is

Xi 0 1 2

P(xi) K 2k 3k

• (a) P(x = 0) + P(x = 1) + P(x = 2) • =1

• K + 2k + 3k = 6k = 1

• K =

• (b) P(x < 2 ) = P(O) + P(1)

• = K + 2k = 3k + =

• P(X ≤ 2) = P(X = 0) + P)x = 1) + P (x = 2)

• = K + 2k + 3k = 6k = 6 × = 1

• P(X ≥ 2) = P(2)

• = 3k = 3 =



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53. Find the mean number of heads in three tosses of a fair coin.

• Solution: S = {H, T}

x(S) = 2

Let A denote the appearance of head on a toss.

A = [h]

n(A) = 1

P = =

q = 1 - p = 1 - =

n = 3, r = 0, 1, 2, 3

P(X = 0) = q3 = =

P(X = 1) = 3q2p = 3 × =

P(X = 2) = 3qp2 = 3

=

P(X = 3) = p3 = =

Xi 0 1 2 3

Pi

Mean = = 0 × + 1 × + 2 × + 3 ×



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= =

54. Two dice are thrown simultaneously. If X denotes the number of sixes, find the

expectation of X.

• Solution: Two disc are thrown simultaneously in equivalent to the several disc thrown 2 times

Let S = {1, 2, 3, 4, 5, 6}

n(S) = 6

Let A denotes the number 6

A = {6} n(A) = 1

P(A) = =

P( ) = 1- =

Now n = 2, r = 0, 1 ,2

P(X = 0) = P( P( = =

P(X = 1) = 2 p(A) P( = 2 × ×

=

P(X = 2) = p(A) P(A)= × =

E(X) = P(xi) = O × + 2 ×

= =

55. Two numbers are selected at random (without replacement) from the first six

positive integers. Let X denote the larger of the two numbers obtained. Find E(X).



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• Solution: Sample space

S =

n(S) = 30

Let X denote the larger of the two numbers obtained

xi fi pi pixi

2 2

3 4

4 6

5 8

6 10

30

∑ xi pi =

E(x) = ∑ pixini = 4

56. Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the

variance and standard deviation of X.

• Solution:

S =

n(s) = 36

Let the sum be as below

A = 2, B = 3, C = 4, D = 5, E = 6, F = 7, G = 8, H = 9, I = 10, J = 11, K = 12



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When A = {11} n(A) = 1, P(A) = , P(B) = , P(C) = , P(D) = ,

P(E) = , P(F) = , P(G) = , P(H) = , P(I) = , P(J) = ,

P(K) = .

xi p(xi) xi P(x)

2

7

3

8

4

9

5

10

6

12

Mean µ = Σ Pini

=

= = 7

Σ Pixi2 =

= =

Variance = Σ Pixi2 – (Σ Pixi)

2

= = 5.83

Standard deviation = = 2.41



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57. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

• Solution:

n(S) = 15 P(A) =

xi fi pi pixi Pixi2

14 2

15 1

16 2

17 3

18 1

19 2

20 3

21 1

• Σ Pixi = Σ Pixi2 =

• Mean = Σ Pixi = = 17.5

• Variance = Σ Pixi2 - (Σ Pixi)

2

• = -

• = 4.78

• Standard deviation = 2.19

58. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A

member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var (X). Choose the correct answer in each of the following:

• Solution:

X pi pi.xi pi.xi2

0

0 0



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1

1 Σ Pipi

Σ Pixi2 =

• E(x) = Mean = Σ pixi

• = = 0.7

• var(x) = Σ pixi2 - (Σ pixi)2

• = - = 0.21

59. A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the

probability of

(i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes?

• Solution: S= {1, 2, 3, 4, 5, 6}

n(S) = 6

A = {1, 3, 5}

n(A) = 3

P = = =

q = 1 – p = 1 - =

n = 6

P(x = r) = (n, r) pr. qn – r

(i) r = 5

P(x = 5) = ((6, 5)) = 6

= =

(ii) r = 5, 6



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n(at least 5 successes)

= P(x = 5) + P(x = 6)

= + = =

(iii) P(at most 5 successes)

= 1 – P(x = 6) = 1 - =

60. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the

probability of two successes.

• Solution: n(s) = 36

A = {11, 22, 33, 44, 55, 66}

x(A) = 6

P = = =

q = 1 – p = 1 - =

n = 4, r = 2

P(x = r), c(p, r) pr . qn- v

n = 4, r = 2

P(x = 2), C(4, 2)

= 6 × 25 × =

61. There are 5% defective items in a large bulk of items. What is the probability that a

sample of 10 items will include not more than one defective item?



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• Solution:

q = =

q = 1 - P = 1 - =

n = 10 r = 0, 1

P(not more than one defective items)

= P(x = 0) + P(x = 1)

= +

= +

62. Five cards are drawn successively with replacement from a well-shuffled deck of 52

cards. What is the probability that

(i) all the five cards are spades?

(ii) only 3 cards are spades?

(iii) none is a spade?

• Solution: S = [52 cards]

n(s) = 52

Let A denotes the favorable events

A= [13 spade]

n(A) = 13

p = = =

q = 1 – p = 1 - =

(i) p = 5 r = 5



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p(x : v) = nCr pr.qn – v

P(x = 5) = 5C5 =

(ii) n = 5 r = 3

P(X = 3) = 5C3

=

(iii) n = 5 y = 0

p(x = 0) =

63. The probability that a bulb produced by a factory will fuse after 150 days of use is

0.05. Find the probability that out of 5 such bulbs

(i) none

(ii) not more than one

(iii) more than one

(iv) at least one

will fuse after 150 days of use.

• Solution: p = 0.05

q = 1 – p = 0.95

P(x = r) ncr prqn – r

i) r = 0

P(x = 0) =

(ii) p(not more than one success)

= P(x = 0) + P(x = 1)



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= + 5C1 (0.05)(0.95)4

=

(iii) P(more than one success)

1 – [P(x = 0) + P(x = 1)]

= 1 -

(iv) P(at least one success)

= 1 – P(x = 0) = 1 -

64. A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are

drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

• Solution: S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

n(S) = 10

Let A represents that the ball is marked with the digit 0.

A = {0}

n(A) = 1

P = q = 1 – p = =

n = 4 r = 0

P(none is marked with digit 0)

= 4C0(p)0(q)4 = =



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65. In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false'. Find the probability that he answers at least 12 questions correctly.

• Solution: S = {H, T}

n(S) = 2

Let A represent a head

A = [H}

n(A) = 1

p = q =

n = 20

r = 12, 13 ……………20

p(x = r) = pcr pr – q n – r

P(at least 12 success)

P(x = 12) +P(X = 13) +……. P(x = 20)

=

66. Suppose X has a binomial distribution 1 B . Show that X = 3 is the most likely outcome. (Hint : P(X = 3) is the maximum among all P(xi), xi = 0,1,2,3,4,5,6)

• Solution:

B

n = 6 p = q =

x1 = 0, 1, 2, 3, 4, 5, 6

P(x = r) = nCr pr qn – r

As p = q



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P(x = r) = nCr px

6C3 is maximum.

=

=

67. On a multiple choice examination with three possible answers for each of the five

questions, what is the probability that a candidate would get four or more correct answers just by guessing?

• Solution:

p =

q = 1 – p = 1 - =

n = 5 r = 4, 5

p(x = r) ncr pr qn – r

P(Four or more success)

= P(x = 4) + p(x = 5)

= 5c4 + 5C5

=

68. A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning

a prize is .What is the probability that he will win a prize (a) at least once (b) exactly

once (c) at least twice?

• Solution:

p =



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q = 1 – p =

n = 50

(a) P(at least one prize)

= 1 – p(x = 0) = 1 -

(b) P(exactly one prize)

p(x = 1) = 50c1

=

(c) P(at least twice)

= 1 – [P(x = 0) + p(x = 1)]

= 1 –

69. Find the probability of getting 5 exactly twice in 7 throws of a die.

• Solution: S = {1, 2, 3, 4, 5, 6}

n (s) = 6

A = [5]

P = =

q = 1 - =

n = 7, r = 2

p(x = r) nCr pr qn – r



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p(x = 2) = 7C2

=

70. Find the probability of throwing at most 2 sixes in 6 throws of a single die.

• Solution: S = {1, 2, 3, 4, 5, 6]

n(S) = 6

Let A represents the favourable event (ie) 6

A = [6]

n(A) = 1

p = q =

n = 6, r = 0, 1, 2

p(x = r) = ncr pr . qn - r

p(x = 0) =

p(x = 1) = 6C1 =

p(x = 2) = 6C2

= 15

p(at most 2 success)

= p(x = 0) + p(x = 1) + p(x = 2)



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=

=

= = 9;

71. It is known that 10% of certain articles manufactured are defective. What is the

probability that in a random sample of 12 such articles, 9 are defective?

• Solution:

p = =

q = 1 – p = 1 - =

n = 12, r = 9

P(x = r) = ncr pr qp – r

P(x = 9) = 12c9

=



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