Probability - Aptitude MCQ Questions and Solutions with Explanations

Total cases = [H, T] - 2
Favourable cases = [H] - 1
So probability of getting head = $$\frac{{1}}{{2}}$$

Number greater than 5 is 6, so only 1 number
Total cases of dice = [1, 2, 3, 4, 5, 6]
So probability = $$\frac{{1}}{{6}}$$

Total number of drink bottles = 6 + 3 + 4 = 13
Let S be the sample space
Then, n(S) = number of ways of taking 3 drink bottles out of 13
Therefore,
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^{13}{C_3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{13 \times 12 \times 11}}{{1 \times 2 \times 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 22 \times 13 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 286 \cr} $$
Let E be the event of taking 3 bottles of the same variety.
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right) = {}^6{C_3} + {}^3{C_3} + {}^4{C_3} \cr & = \frac{{6 \times 5 \times 4}}{{1 \times 2 \times 3}} + 1 + \frac{{4 \times 3 \times 2}}{{1 \times 2 \times 3}} \cr & = 20 + 1 + 4 \cr & = 25 \cr} $$
The probability of taking 3 bottles of the same variety $$ = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{25}}{{858}}$$
Then, the probability of taking 3 bottles are not of the same variety
$$\eqalign{ & = 1 - \frac{{25}}{{286}} \cr & = \frac{{261}}{{286}} \cr} $$

Total cases = [H, T] - 2
Favourable cases = [T] -1
So probability of getting tails = $$\frac{{1}}{{2}}$$

In a simultaneous throw of two dice, we have n(s) = 6 × 6 = 36
Let E = event of getting two numbers whose sum is at most 5
Then E = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}
therefore, n(E) = 10
And p(E) = p( getting two numbers whose sum is at most 5)
$$\eqalign{ & {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{{36}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{{18}} \cr} $$
Hence the answer is $$\frac{{5}}{{18}}$$

Total number of students = 20
Let S be the sample space
Then, n(S) = number of ways of three scored first mark
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^{20}{C_3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20 \times 19 \times 18}}{{2 \times 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20 \times 19 \times 3 \cr} $$
Let, E be the event of 1 girl and 2 boys
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right) = {}^8{C_1} \times {}^{12}{C_2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 12 \times 11}}{{1 \times 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \times 6 \times 11 \cr} $$
Now, the required probability
$$\eqalign{ & = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & = \frac{{8 \times 6 \times 11}}{{20 \times 19 \times 3}} \cr & = \frac{{44}}{{95}} \cr} $$

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green = event that the ball drawn is red.
Therefore, n(E) = 8
P(E) = $$\frac{{8}}{{21}}$$

In a simultaneous throw of two dice, we have n(s) = 6 × 6 = 36
Let E = event of getting two number whose sum is at least 10
Then E = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
therefore, n(E) = 6
And p(E) = p(getting two numbers whose sum is at least 10)
$$\eqalign{ & {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{{36}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{6} \cr} $$
Hence the answer is $$\frac{{1}}{{6}}$$

Let S be the sample space
Total number of students in the class = 12 boys + 16 girls = 28
Then, n(S) = 28
Let E be the event of calling one of them by enroll number
Given that, number of girls = 16
Then, n(E) = 16
The probability that the one called is a girl
$$\eqalign{ & = \frac{{{\text{n}}\left( {\text{S}} \right)}}{{{\text{n}}\left( {\text{E}} \right)}} \cr & = \frac{{16}}{{28}} \cr & = \frac{4}{7} \cr} $$

Sample space (All [possible outcomes)S=(HH, HT, TH, TT)
Event(required outcomes) = (TT, HT, TH)
$$\eqalign{ & {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{4} \cr} $$