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51. A basket contains 6 blue, 2 red, 4 green, and 3 yellow balls. If three balls are picked up at random, what is the probability that none is yellow?

A. $\frac{3}{455}$

B. $\frac{1}{5}$

C. $\frac{4}{5}$

D. $\frac{44}{91}$

Discuss

Solution:

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 3 non-yellow balls
Then, n(E) =

^{12} C_{3}

= \frac{12 \times 11 \times 10}{3 \times 2 \times 1}

= 220
Also, n(S) =

^{15} C_{3}

= \frac{15 \times 14 \times 13}{3 \times 2 \times 1}

= 455

∴ P (E) = \frac{n (E)}{n (S)} = \frac{220}{455} = \frac{44}{91}

52. The probability that a card drawn from a pack of 52 cards will be a diamond or a king is -

A. $\frac{2}{13}$

B. $\frac{4}{13}$

C. $\frac{1}{13}$

D. $\frac{1}{52}$

Discuss

Solution:

Here, n(S) = 52
There are 13 cards of diamond (including one king) and there are three more kings.
Let E = event of getting a diamond or a king
Then, n(E) = (13 + 3) = 16

∴ P (E) = \frac{n (E)}{n (S)} = \frac{16}{52} = \frac{4}{13}

53. What is the probability of getting a sum 9 from two throws of a dice ?

A. $\frac{1}{6}$

B. $\frac{1}{8}$

C. $\frac{1}{9}$

D. $\frac{1}{12}$

Discuss

Solution:

In two throws of dice, n(S) = (6 × 6) = 36
Let E = event of getting a sum 9 = [(3, 6), (4, 5), (5, 4), (6, 3)]

∴ P (E) = \frac{n (E)}{n (S)} = \frac{4}{36} = \frac{1}{9}

54. Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is-

A. $\frac{3}{20}$

B. $\frac{29}{34}$

C. $\frac{47}{100}$

D. $\frac{13}{102}$

Discuss

Solution:

Let S be the sample space.
Then, n(S) =

^{52} C_{2}

= \frac{(52 \times 51)}{(2 \times 1)}

= 1326
Let E = event of getting 1 spade and 1 heart.
∴ n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13

= (^{13} C_{1} \times^{13} C_{1})

= (13 × 13) = 169

∴ P (E) = \frac{n (E)}{n (S)} = \frac{169}{1326} = \frac{13}{102}

55. A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least 1 woman ?

A. $\frac{1}{10}$

B. $\frac{9}{20}$

C. $\frac{1}{20}$

D. $\frac{9}{10}$

Discuss

Solution:

Total number of persons = (3 + 2) = 5

∴ n (S) =^{5} C_{3} =^{5} C_{2}

= \frac{5 \times 4}{2 \times 1}

= 10
Let E be the event of selecting 3 members having at least 1 women
Then, n(E) = n [(1 women and 2 men ) or (2 women and 1 man)]
= n (1 woman and 2 men) + n (2 women and 1 man)

\begin{aligned} = (^{2} C_{1} \times^{3} C_{2}) + (^{2} C_{2} \times^{3} C_{1}) \\ = (^{2} C_{1} \times^{3} C_{1}) + (1 \times^{3} C_{1}) \\ = (2 \times 3) + (1 \times 3) \\ = (6 + 3) \\ = 9 \end{aligned}

∴ P (E) = \frac{n (E)}{n (S)} = \frac{9}{10}

56. A speaks truth in 75% cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, in narrating the same incident ?

A. 5%

B. 15%

C. 35%

D. 45%

Discuss

Solution:

Let

E_{1}

= event that A speaks the truth
And

E_{2}

= event that B speaks the truth
Then,

\begin{aligned} P (E_{1}) = \frac{75}{100} = \frac{3}{4}, \\ P (E_{2}) = \frac{80}{100} = \frac{4}{5}, \\ P ({\bar{E}}_{1}) = (1 - \frac{3}{4}) = \frac{1}{4}, \\ P ({\bar{E}}_{2}) = (1 - \frac{4}{5}) = \frac{1}{5} \end{aligned}

P (A and B contradict each other)
= P [(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)]

\begin{aligned} = P [(E_{1} \cap {\bar{E}}_{2}) o r ({\bar{E}}_{1} \cap E_{2})] \\ = P [(E_{1} \cap {\bar{E}}_{2}) + ({\bar{E}}_{1} \cap E_{2})] \\ = P (E_{1}) . P ({\bar{E}}_{2}) + P ({\bar{E}}_{1}) . P (E_{2}) \\ = (\frac{3}{4} \times \frac{1}{5}) + (\frac{1}{4} \times \frac{4}{5}) \\ = (\frac{3}{20} + \frac{1}{5}) \\ = \frac{7}{20} \\ = (\frac{7}{20} \times 100) % \\ = 35 % \end{aligned}

57. A basket contains 4 red, 5 blue and 3 green marbles. If 2 marbles are drawn at random from the basket, What is the probability that both are red ?

A. $\frac{3}{7}$

B. $\frac{1}{2}$

C. $\frac{1}{11}$

D. $\frac{1}{6}$

Discuss

Solution:

Total number of balls = (4 + 5 + 3) = 12
Let E be the event of drawing 2 red balls.
Then, n (E) =

^{4} C_{2} = \frac{4 \times 3}{2 \times 1}

= 6
Also n (S) =

^{12} C_{2} = \frac{12 \times 11}{2 \times 1}

= 66

∴ P (E) = \frac{n (E)}{n (S)} = \frac{6}{66} = \frac{1}{11}

58. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize ?

A. $\frac{1}{10}$

B. $\frac{2}{5}$

C. $\frac{2}{7}$

D. $\frac{5}{7}$

Discuss

Solution:

P (getting a prize)

\begin{aligned} = \frac{10}{(10 + 25)} \\ = \frac{10}{35} \\ = \frac{2}{7} \end{aligned}

59. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green ?

A. $\frac{2}{5}$

B. $\frac{3}{4}$

C. $\frac{7}{19}$

D. $\frac{8}{21}$

Discuss

Solution:

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green
= Even that the ball drawn is red
∴ n(E) = 8
P(E) =

\frac{8}{21}

60. Dev can hit a target 3 times in 6 shots pawan can hit the target 2 times in 6 shots and Lakhan can hit the target 4 times in 4 shots. What is the probability that at least 2 shots hit the target -

A. $\frac{2}{3}$

B. $\frac{1}{3}$

C. $\frac{1}{2}$

D. None of these

Discuss

Solution:

Probability of hitting the target:
Dev can hit target ⇒

\frac{3}{6}

\frac{1}{2}

Lakhan can hit target =

\frac{4}{4}

= 1
Pawan can hit target =

\frac{2}{6}

\frac{1}{3}

Required probability that at least 2 shorts hit target

\begin{aligned} = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} \\ = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} \\ = \frac{4}{6} \\ = \frac{2}{3} \end{aligned}

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