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21. A bag contains 4 red balls, 6 blue balls and 8 pink balls. One ball is drawn at random and replaced with 3 pink balls. A probability that the first ball drawn was either red or blue in colour and the second drawn was pink in colour ?

A. $\frac{12}{21}$

B. $\frac{13}{17}$

C. $\frac{11}{30}$

D. $\frac{13}{18}$

Discuss

Solution:

Number of Red balls = 4
Number of Blue balls = 6
Number of Pink balls = 8
Total number of balls = 4 + 6 + 8 = 18
Required probability

\begin{aligned} = \frac{4}{18} \times \frac{11}{20} + \frac{6}{18} \times \frac{11}{20} \\ = \frac{11}{20} [\frac{4}{18} + \frac{6}{18}] \\ = \frac{11}{20} \times \frac{10}{18} \\ = \frac{11}{36} \end{aligned}

22. A bag contains 3 blue, 2 green and 5 red balls. If four balls are picked at random, what is the probability that two are green and two are blue?

A. $\frac{1}{18}$

B. $\frac{1}{70}$

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

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

Discuss

Solution:

Number of blue balls = 3
Number of green balls = 2
Numbers of red balls = 5
Total balls in the bag = 3 + 2 + 5 = 10
Total possible outcomes = Selection of 4 balls out of 10 balls

=^{10} C_{4} = \frac{10!}{4! \times (10 - 4)!}

= \frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}

= 210
Favorable outcomes = (selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls)

\begin{aligned} =^{2} C_{2} \times^{3} C_{2} \\ = 1 \times 3 \\ = 3 \end{aligned}

∴ Required probability

= \frac{Favorable outcomes}{Total possible outcomes}

\begin{aligned} = \frac{3}{210} \\ = \frac{1}{70} \end{aligned}

23. A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If 5 balls are picked up at random, what is the probability that at least one is blue?

A. $\frac{137}{143}$

B. $\frac{18}{455}$

C. $\frac{9}{91}$

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

Discuss

Solution:

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 5 balls out of 9 non-blue balls.

=^{9} C_{5} =^{9} C_{(9 - 5)} =^{9} C_{4}

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

= 126
And,
n(S) =

^{15} C_{5} =

\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}

= 3003

∴ P (E) = \frac{n (E)}{n (S)} =

\frac{126}{3003} = \frac{6}{143}

∴ Required Probability

\begin{aligned} = (1 - \frac{6}{143}) \\ = \frac{137}{143} \end{aligned}

24. Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

A. $\frac{1}{4}$

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

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

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

Discuss

Solution:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}

∴ P (E) = \frac{n (E)}{n (S)} = \frac{4}{8} = \frac{1}{2}

25. In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11?

A. $\frac{1}{4}$

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

C. $\frac{7}{12}$

D. $\frac{5}{36}$

Discuss

Solution:

In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36
Let E = event of getting a total of 10 or 11
= [(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)]

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

26. From a pack of 52 cards, one card is drawn at random. What is the probability that the card drawn is a ten or a spade?

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

B. $\frac{1}{4}$

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

D. $\frac{1}{26}$

Discuss

Solution:

Hence, n (S) = 52
There are 13 spades (including one ten) and there are 3 more ten
Let E = event of getting a ten or a spade
Then, n (E) = (13 + 3) = 16

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

27. A box contains 4 red, 5 green and 6 white balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green ?

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

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

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

D. $\frac{7}{15}$

Discuss

Solution:

Total number of balls = (4 + 5 + 6) = 15
P (drawing a red ball or a green ball) = P (red) + P (green)

\begin{aligned} = (\frac{4}{15} + \frac{5}{15}) \\ = \frac{9}{15} \\ = \frac{3}{5} \end{aligned}

28. Three unbiased coins are tossed. What is the probability of getting at most two heads?

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

B. $\frac{1}{4}$

C. $\frac{3}{8}$

D. $\frac{7}{8}$

Discuss

Solution:

Here S = [TTT, TTH, THT, HTT, THH, HTH, HHT, HHH]
Let E = event of getting at most two heads
Then, E = {TTT, TTH, THT, HTT, THH, HTH, HHT}

∴ P (E) = \frac{n (E)}{n (S)} = \frac{7}{8}

29. A bag contains 6 red balls 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag. One after another what is the probability that the first ball is red and second ball is yellow?

A. $\frac{1}{14}$

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

C. $\frac{1}{7}$

D. $\frac{3}{14}$

Discuss

Solution:

Number of red balls = 6
Number of yellow balls = 11
Number of pink balls = 5
Total number of balls = 6 + 11 + 5 = 22
Total possible outcomes

n (E) =^{22} C_{2} = \frac{22!}{2! (22 - 2)!}

= \frac{22!}{2! \times 20!}

= \frac{22 \times 21}{2 \times 1}

=

231
Number of favourable outcomes

n (S) =^{6} C_{1} \times^{11} C_{1}

= 6 × 11 = 66
Required probability =

\frac{n (E)}{n (S)}

= \frac{66}{231}

= \frac{1}{7}

30. A man and his wife appear in an interview for two vacancies in the same post. The probability two of husband’s selection is $\frac{1}{7}$ and the probability of wife’s selection is $\frac{1}{5}$ . What is the probability that only one of them is selected?

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

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

C. $\frac{4}{7}$

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

Discuss

Solution:

Let

E_{1}

= event that the husband is selected and

E_{2}

= event that the wife is selected.
Then,

P (E_{1}) = \frac{1}{7}

and

P (E_{2}) = \frac{1}{5}

∴

P ({\bar{E}}_{1}) = (1 - \frac{1}{7})

= \frac{6}{7}

and

P ({\bar{E}}_{2}) =

(1 - \frac{1}{5}) = \frac{4}{5}

∴ Required probability = P [( A and not B) or (B and not A)]
= P

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

= P

(E_{1} \cap {\bar{E}}_{2}) + P (E_{2} \cap {\bar{E}}_{1})

P (E_{1}) . P ({\bar{E}}_{2}) + P (E_{2}) . P ({\bar{E}}_{1})

\begin{aligned} = (\frac{1}{7} \times \frac{4}{5}) + (\frac{1}{5} \times \frac{6}{7}) \\ = \frac{10}{35} \\ = \frac{2}{7} \end{aligned}

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