Probability - Aptitude MCQ Questions and Solutions with Explanations

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

Probability of hitting the target:
Dev can hit target ⇒ $$\frac{{3}}{{6}}$$ =$$\frac{{1}}{{2}}$$
Lakhan can hit target =$$\frac{{4}}{{4}}$$  = 1
Pawan can hit target = $$\frac{{2}}{{6}}$$  = $$\frac{{1}}{{3}}$$
Required probability that at least 2 shorts hit target
$$\eqalign{ & = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} \cr & = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} \cr & = \frac{4}{6} \cr & = \frac{2}{3} \cr} $$

Total number of balls = (6 + 8) = 14
Number of white balls = 8
P (drawing a white ball) = $$\frac{{8}}{{14}}$$  = $$\frac{{4}}{{7}}$$

Number of red balls = 4
Number of yellow balls = 5
Number of pink balls = 6
Total balls = 4 + 5 + 6 = 15
Total possible outcomes = selection of 2 balls out of 15 balls
= $${}^{15}\mathop C\nolimits_2 $$  
=$$\frac{{15!}}{{2!(15 - 2)!}}$$
=$$\frac{{15!}}{{2! × 13!}}$$
=$$\frac{{15 × 14}}{{1 × 2}}$$
=105
Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.
$$ = {}^{10}\mathop C\nolimits_2 $$
= $$\frac{{10!}}{{2!(10 - 2)!}}$$
= $$\frac{{10!}}{{2! × 8!}}$$
= $$\frac{{10 × 9}}{{1 × 2}}$$
= 45
∴ Required probability = $$\frac{{45}}{{105}}$$ = $$\frac{{3}}{{7}}$$

In a simultaneous throw of dice, n (S) = (6 × 6) = 36
Let E = event of getting a doublet
= [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
∴ P(E) = $$\frac{{n (E)}}{{n (S)}}$$ = $$\frac{{6}}{{36}}$$ = $$\frac{{1}}{{6}}$$

P(Both are red),
$$\eqalign{ & = \frac{{^6{C_2}}}{{^{13}{C_2}}} \cr & = \frac{5}{{26}} \cr} $$

Probability of getting a tail when a single coin is tossed $$ = \frac{1}{2}$$
Probability of getting 4 when a die is thrown $$ = \frac{1}{6}$$
Required probability,
$$\eqalign{ & = \frac{1}{2} \times \frac{1}{6} \cr & = \frac{1}{{12}} \cr} $$

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
$$\eqalign{ & = P(A) + P(B) \cr & = \frac{{^{12}{C_1}{ \times ^1}{C_1}}}{{^{52}{C_2}}} + \frac{{^{13}{C_1}{ \times ^3}{C_1}}}{{^{52}{C_2}}} \cr & = \left( {\frac{{2 \times \left( {12 \times 1} \right)}}{{52 \times 51}}} \right) + \left( {\frac{{2\left( {13 \times 3} \right)}}{{52 \times 51}}} \right) \cr & = \left( {\frac{{24 + 78}}{{52 \times 51}}} \right) \cr & = \frac{1}{{26}} \cr} $$

n(S)= number of ways of sitting 12 persons at round table:
= (12 - 1)! = 11!
Since two persons will be always together, then number of persons:
= 10 + 1 = 11
So, 11 persons will be seated in (11 - 1)! = 10! ways at round table and 2 particular persons will be seated in 2! ways.
n(A) = The number of ways in which two persons always sit together = 10! × 2
$$\eqalign{ & P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}} \cr & = \frac{{10!\, \times 2!}}{{11!}} \cr & = \frac{2}{{11}} \cr} $$

Let the probability of winning the race is denoted by P(person)
$$\eqalign{ & P(P) = \frac{1}{4}, \cr & P(Q) = \frac{1}{5}, \cr & P(R) = \frac{1}{6}, \cr & P(S) = \frac{1}{7} \cr} $$
All the events are mutually exclusive (since if one of them wins then other would lose as pointed out by rahul) hence,
Required probability:
$$ = P(P) + P(Q) + P(R)$$     $$ + P(S)$$
$$\eqalign{ & = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \cr & = \frac{{319}}{{420}} \cr} $$