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41. An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 Is green, 2 are blue and 1 is red ?

A. $\frac{13}{35}$

B. $\frac{24}{455}$

C. $\frac{11}{15}$

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

Discuss

Solution:

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 1 green, 2 blue and 1 red marble.
Then,
n (E) =

^{2} C_{1} \times^{4} C_{2} \times^{6} C_{1}

= 2 \times \frac{4 \times 3}{2 \times 1} \times 6

= 72
And, n (S) =

^{15} C_{4} =

\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}

= 1365

∴ P (E) = \frac{n (E)}{n (S)} = \frac{72}{1365}

= \frac{24}{455}

42. An urn contains 2 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.

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

B. $\frac{10}{21}$

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

D. $\frac{11}{21}$

Discuss

Solution:

Total number of balls = (2 + 3 + 2) = 7
Let, E be the event of drawing 2 non-blue balls.
Then, n (E) =

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

= 10
And, n (S) =

^{7} C_{2} = \frac{7 \times 6}{2 \times 1}

= 21

∴ P (E) = \frac{n (E)}{n (S)} = \frac{10}{21}

43. Two dice are tossed. The probability that the total score is a prime number is-

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

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

C. $\frac{5}{12}$

D. $\frac{7}{9}$

Discuss

Solution:

Clearly, n (S) = (6 × 6) = 36
Let E be the event that the sum is a prime number. Then, n (E) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3,), (5, 2), (5, 6), (6, 1), (6, 5)}
∴ n (E) = 15

∴ P (E) = \frac{n (E)}{n (S)} = \frac{15}{36} = \frac{5}{12}

44. A box contains 10 black and 10 white balls. What is the probability of drawing 2 balls of the same colour ?

A. $\frac{9}{19}$

B. $\frac{9}{38}$

C. $\frac{10}{19}$

D. $\frac{5}{19}$

Discuss

Solution:

Total number of balls = (10 + 10) = 20
Let E be the event of drawing 2 balls of the same colour.
n (E) = number of ways of drawing 2 black balls or 2 white balls.
n (E) =

(^{10} C_{2} \times^{10} C_{2})

= 2 \times^{10} C_{2}

= 2 \times \frac{10 \times 9}{2 \times 1}

= 90
n (S) = number of ways of drawing 2 balls out of 20 balls

=^{20} C_{2} = \frac{20 \times 19}{2 \times 1}

= 190

∴ P (E) = \frac{n (E)}{n (S)} = \frac{90}{190} = \frac{9}{19}

45. A bag contains 10 mangoes out of which 4 are rotten, two mangoes are taken out together. If one of them is found to be good, the probability that other also good is-

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

B. $\frac{8}{15}$

C. $\frac{5}{18}$

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

Discuss

Solution:

Out of mangoes, 4 mangoes are rotten
∴ Required probability

\begin{aligned} = \frac{^{6} C_{2}}{^{10} C_{2}} \\ = \frac{\frac{6!}{2! (6 - 2)!}}{\frac{10!}{2! (10 - 2)!}} \\ = \frac{\frac{6!}{2! 4!}}{\frac{10!}{2! \times 8!}} \\ = \frac{\frac{6 \times 5}{1 \times 2}}{\frac{10 \times 9}{1 \times 2}} \\ = \frac{6 \times 5}{10 \times 9} \\ = \frac{1}{3} \end{aligned}

46. A speaks truth in 60% cases B speaks truth in 70% cases. The probability that they will way say the same thing while describing a single event, is-

A. 0.54

B. 0.56

C. 0.68

D. 0.94

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{60}{100} = \frac{3}{5}, \\ P (E_{2}) = \frac{70}{100} = \frac{7}{10}, \\ P ({\bar{E}}_{1}) = (1 - \frac{3}{5}) = \frac{2}{5}, \\ P ({\bar{E}}_{2}) = (1 - \frac{7}{10}) = \frac{3}{10} \end{aligned}

P (A and B say the same thing) = P [(A speaks the truth and B speaks the truth) or (A tells a lie and B tells a lie)]

=

P [

(E_{1} \cap E_{2})

({\bar{E}}_{1} \cap {\bar{E}}_{2})]

=

(E_{1} \cap E_{2})

({\bar{E}}_{1} \cap {\bar{E}}_{2})

=

P (

E_{1}

). P (

E_{2}

) + P (

{\bar{E}}_{1}

) . P (

{\bar{E}}_{2}

)

\begin{aligned} = (\frac{3}{5} \times \frac{7}{10}) + (\frac{2}{5} \times \frac{3}{10}) \\ = \frac{27}{50} \\ = 0.54 \end{aligned}

47. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that the selected students are 2 boys and 1 girls, is:

A. $\frac{21}{46}$

B. $\frac{25}{117}$

C. $\frac{1}{50}$

D. $\frac{3}{25}$

Discuss

Solution:

Let S be the sample space and let E be the event of selecting 2 boys and 1 girl.
Then, n(S) = number of ways of selecting 3 students out of 25
=

^{25} C_{3} =

\frac{25 \times 24 \times 23}{3 \times 2 \times 1}

= 2300
And, n(E) =

(^{15} C_{2} \times^{10} C_{1})

= (\frac{15 \times 14}{2 \times 1} \times 10)

= 1050

∴ P (E) = \frac{n (E)}{n (S)} = \frac{1050}{2300} = \frac{21}{46}

48. A basket contains 6 blue, 2 red,4 green and 3 yellow balls. If four balls are picked up at random, what is the probability that 2 are red 2 are green?

A. $\frac{4}{15}$

B. $\frac{5}{27}$

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

D. $\frac{2}{445}$

Discuss

Solution:

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 4 balls such that 2 are red and 2 are green.
Then, n(E) =

(^{2} C_{2} \times^{4} C_{2})

= (1 \times \frac{4 \times 3}{2 \times 1})

= 6
And, n(S) =

^{15} C_{4} =

\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}

= 1365

∴ P (E) = \frac{n (E)}{n (S)} = \frac{6}{1365} = \frac{2}{455}

49. In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, What is the probability that he has offered English or Hindi ?

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

B. $\frac{3}{5}$

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

D. $\frac{3}{10}$

Discuss

Solution:

\begin{aligned} P (E) = \frac{30}{100} = \frac{3}{10}; \\ P (H) = \frac{20}{100} = \frac{1}{5} and \\ P (E \cap H) = \frac{10}{100} = \frac{1}{10} \end{aligned}

P (E or H) =

P (E \cup H)

= P(E) + P(H) -

P (E \cup H)

\begin{aligned} = (\frac{3}{10} + \frac{1}{5} - \frac{1}{10}) \\ = \frac{4}{10} \\ = \frac{2}{5} \end{aligned}

50. Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children, is-

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

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

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

D. $\frac{10}{21}$

Discuss

Solution:

n(S) = number of ways of choosing 4 persons out of 9

=^{9} C_{4}

= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}

= 126
n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E)

= (^{4} C_{2} \times^{5} C_{2})

= (\frac{4 \times 3}{2 \times 1} \times \frac{5 \times 4}{2 \times 1})

= 60

∴ P (E) = \frac{n (E)}{n (S)} = \frac{60}{126} = \frac{10}{21}

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