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81. A bag contains 21 toys numbered 1 to 21. A toy is drawn and then another toy is drawn without replacement.
Find the probability that both toys will show even numbers.

A. $\frac{5}{21}$

B. $\frac{9}{42}$

C. $\frac{11}{42}$

D. $\frac{4}{21}$

Discuss

Solution:

The probability that first toy shows the even number

= \frac{10}{21}

Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows the even number

= \frac{9}{20}

Required probability,

\begin{aligned} = (\frac{10}{21}) \times (\frac{9}{20}) \\ = \frac{9}{42} \end{aligned}

82. Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice.

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

B. $\frac{11}{36}$

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

D. $\frac{13}{36}$

Discuss

Solution:

Let X be required events and S be the sample space
then X = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (6, 2), (3, 4), (6, 4), (3, 6)}
n(X) = 11, n(S) = 36
Hence, required probability

\begin{aligned} = \frac{n (X)}{n (S)} \\ = \frac{11}{36} \end{aligned}

83. The probability of success of three students X,Y and Z in the one examination are $\frac{1}{5}$ , $\frac{1}{4}$ and $\frac{1}{3}$ respectively. Find the probability of success of at least two.

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

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

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

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

Discuss

Solution:

P (X) = \frac{1}{5},

P (Y) = \frac{1}{4},

P (Z) = \frac{1}{3}

Required probability:

= [P (A) P (B) {1 - P (C)}]

+

[{1 - P (A)} P (B) P (C)] +

[P (A) P (C) {1 - P (B)}] +

[P (A) P (B) P (C)]

= [\frac{1}{4} \times \frac{1}{3} \times \frac{4}{5}] +

[\frac{3}{4} \times \frac{1}{3} \times \frac{1}{5}] +

[\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}] +

[\frac{1}{4} \times \frac{1}{3} \times \frac{1}{5}]

\begin{aligned} = \frac{4}{60} + \frac{3}{60} + \frac{2}{60} + \frac{1}{60} \\ = \frac{10}{60} \\ = \frac{1}{6} \end{aligned}

84. A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement.
What is the probability that first is white and second is black?

A. $\frac{18}{145}$

B. $\frac{18}{29}$

C. $\frac{36}{135}$

D. $\frac{36}{145}$

Discuss

Solution:

The probability that first ball is white:

\begin{aligned} = \frac{^{12} C_{1}}{^{30} C_{1}} \\ = \frac{12}{30} \\ = \frac{2}{5} \end{aligned}

Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:

\begin{aligned} = \frac{^{18} C_{1}}{^{29} C_{1}} \\ = \frac{18}{29} \end{aligned}

Required probability,

\begin{aligned} = (\frac{2}{5}) \times (\frac{18}{29}) \\ = \frac{36}{145} \end{aligned}

85. There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

A. $\frac{6}{7}$

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

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

D. $\frac{5}{9}$

Discuss

Solution:

Total cases of checking in the hotels =

4^{3}

ways.
Cases, when 3 men are checking in different hotels = 4 × 3 × 2 = 24 ways.
Required probability:

\begin{aligned} = \frac{24}{4^{3}} \\ = \frac{3}{8} \end{aligned}

86. A speaks truth in 75% of cases and B in 80% of cases. In what percent of cases are they likely to contradict each other in narrating the same event?

A. 35%

B. 5%

C. 45%

D. 22.5%

Discuss

Solution:

Different possible cases of contradiction,
A speaks truth and B does not speaks truth.
Or, A does not speak truth and B speaks truth.

\begin{aligned} = (\frac{3}{4} \times \frac{1}{5}) + (\frac{1}{4} \times \frac{4}{5}) \\ = \frac{3}{20} + \frac{4}{20} \\ = \frac{7}{20} \\ = 35 % \end{aligned}

87. Four dice are thrown simultaneously. Find the probability that all of them show the same face.

A. $\frac{1}{216}$

B. $\frac{1}{36}$

C. $\frac{4}{216}$

D. $\frac{3}{216}$

Discuss

Solution:

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

\begin{aligned} = 6 \times 6 \times 6 \times 6 = 6^{4} \\ n (S) = 6^{4} \end{aligned}

Let X be the event that all dice show the same face.
X = {(1, 1, 1, 1), (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4), (5, 5, 5, 5), (6, 6, 6, 6)}
n(X) = 6
Hence required probability,

\begin{aligned} = \frac{n (X)}{n (S)} = \frac{6}{6^{4}} \\ = \frac{1}{216} \end{aligned}

88. The odds against an event are 5 : 3 and the odds in favour of another independent event are 7 : 5. Find the probability that at least one of the two events will occur.

A. $\frac{69}{96}$

B. $\frac{52}{96}$

C. $\frac{71}{96}$

D. $\frac{13}{96}$

Discuss

Solution:

Let probability of the first event taking place be A and probability of the second event taking place be B.
Then

\begin{aligned} P (A) = \frac{3}{5 + 3} \\ = \frac{3}{8} \\ P (B) = \frac{7}{7 + 5} \\ = \frac{7}{12} \end{aligned}

The required event can be defined as that A takes place and B does not take place (A or B takes place and A does not take place or A takes place and B takes place.)

= [P (A) {1 - P (B)}] +

[P (B) {1 - P (A)}] +

[P (A) P (B)]

= [\frac{3}{8} \times \frac{5}{12}] +

[\frac{5}{8} \times \frac{7}{12}] +

[\frac{3}{8} \times \frac{7}{12}]

\begin{aligned} = \frac{15 + 35 + 21}{96} \\ = \frac{71}{96} \end{aligned}

89. If x is chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

A. $\frac{5}{6}$

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

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

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

Discuss

Solution:

S = {(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7)}
Total element n(S) = 12
xy will be even when even x or y or both will be even.
Events of x, y being even is E.
E = {(1, 6), (2, 5), (2, 6), (2, 7), (3, 6), (4, 5), (4, 6),(4, 7)}
n(E) = 8
So, Probability

\begin{aligned} P = \frac{n (E)}{n (S)} \\ P = \frac{8}{12} \\ P = \frac{2}{3} \end{aligned}

90. A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that $| X | < 2$

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

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

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

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

Discuss

Solution:

| X |

can take 7 values.
To get

| X | < 2

(i.e., -2 < x < + 2) take X = {-1, 0, 1}

P (| X | < 2) =

\frac{Favourable Cases}{Total Cases}

= \frac{3}{7}

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