Probability - Aptitude MCQ Questions and Solutions with Explanations

Given, A be the event that X is selected and B is the event that Y is selected.
$$P(A) = \frac{1}{7},\,P(B) = \frac{2}{9}$$
Let C be the event that both are selected.
P(C) = P(A) × P(B) as A and B are independent events:
$$\eqalign{ & = \left( {\frac{1}{7}} \right) \times \left( {\frac{2}{9}} \right) \cr & = \frac{2}{{63}} \cr} $$

Four cards can be selected from 52 cards in $$^{52}{C_4}$$ ways.
Now, there are four suits, e.g. club, spade, heart and diamond each of 13 cards.
So total number of ways of getting all the four cards of the same suit:
$$\eqalign{ & { \Rightarrow ^{13}}{C_4}{ + ^{13}}{C_4}{ + ^{13}}{C_4}{ + ^{13}}{C_4} \cr & = 4{ \times ^{13}}{C_4} \cr} $$
So required probability,
$$\eqalign{ & = \frac{{4{ \times ^{13}}{C_4}}}{{^{52}{C_4}}} \cr & = \frac{{44}}{{4165}} \cr} $$

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,
$$\eqalign{ & = \left( {\frac{{10}}{{21}}} \right) \times \left( {\frac{9}{{20}}} \right) \cr & = \frac{9}{{42}} \cr} $$

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
$$\eqalign{ & = \frac{{n(X)}}{{n(S)}} \cr & = \frac{{11}}{{36}} \cr} $$

$$P(X) = \frac{1}{5},$$   $$P(Y) = \frac{1}{4},$$   $$P(Z) = \frac{1}{3}$$
Required probability:
$$ = \left[ {P\left( A \right)P\left( B \right)\left\{ {1 - P\left( C \right)} \right\}} \right]$$     $$ + $$ $$\left[ {\left\{ {1 - P\left( A \right)} \right\}P\left( B \right)P\left( C \right)} \right] + $$     $$\left[ {P\left( A \right)P\left( C \right)\left\{ {1 - P\left( B \right)} \right\}} \right] + $$     $$\left[ {P\left( A \right)P\left( B \right)P\left( C \right)} \right]$$
$$ = \left[ {\frac{1}{4} \times \frac{1}{3} \times \frac{4}{5}} \right] + $$   $$\left[ {\frac{3}{4} \times \frac{1}{3} \times \frac{1}{5}} \right] + $$   $$\left[ {\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}} \right] + $$   $$\left[ {\frac{1}{4} \times \frac{1}{3} \times \frac{1}{5}} \right]$$
$$\eqalign{ & = \frac{4}{{60}} + \frac{3}{{60}} + \frac{2}{{60}} + \frac{1}{{60}} \cr & = \frac{{10}}{{60}} \cr & = \frac{1}{6} \cr} $$

The probability that first ball is white:
$$\eqalign{ & = \frac{{^{12}{C_1}}}{{^{30}{C_1}}} \cr & = \frac{{12}}{{30}} \cr & = \frac{2}{5} \cr} $$
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
$$\eqalign{ & = \frac{{^{18}{C_1}}}{{^{29}{C_1}}} \cr & = \frac{{18}}{{29}} \cr} $$
Required probability,
$$\eqalign{ & = \left( {\frac{2}{5}} \right) \times \left( {\frac{{18}}{{29}}} \right) \cr & = \frac{{36}}{{145}} \cr} $$

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:
$$\eqalign{ & = \frac{{24}}{{{4^3}}} \cr & = \frac{3}{8} \cr} $$

Different possible cases of contradiction,
A speaks truth and B does not speaks truth.
Or, A does not speak truth and B speaks truth.
$$\eqalign{ & = \left( {\frac{3}{4} \times \frac{1}{5}} \right) + \left( {\frac{1}{4} \times \frac{4}{5}} \right) \cr & = \frac{3}{{20}} + \frac{4}{{20}} \cr & = \frac{7}{{20}} \cr & = 35\% \cr} $$

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
$$\eqalign{ & = 6 \times 6 \times 6 \times 6 = {6^4} \cr & n(S) = {6^4} \cr} $$
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,
$$\eqalign{ & = \frac{{n(X)}}{{n(S)}} = \frac{6}{{{6^4}}} \cr & = \frac{1}{{216}} \cr} $$

Let probability of the first event taking place be A and probability of the second event taking place be B.
Then
$$\eqalign{ & P\left( A \right) = \frac{3}{{5 + 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{8} \cr & P\left( B \right) = \frac{7}{{7 + 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{7}{{12}} \cr} $$
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.)
$$ = \left[ {P\left( A \right)\left\{ {1 - P\left( B \right)} \right\}} \right] + $$     $$\left[ {P\left( B \right)\left\{ {1 - P\left( A \right)} \right\}} \right] + $$     $$\left[ {P\left( A \right)P\left( B \right)} \right]$$
$$ = \left[ {\frac{3}{8} \times \frac{5}{{12}}} \right] + $$   $$\left[ {\frac{5}{8} \times \frac{7}{{12}}} \right] + $$   $$\left[ {\frac{3}{8} \times \frac{7}{{12}}} \right]$$
$$\eqalign{ & = \frac{{15 + 35 + 21}}{{96}} \cr & = \frac{{71}}{{96}} \cr} $$