Probability - Aptitude MCQ Questions and Solutions with Explanations

A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Saturday, Sunday}
Required Probability = $$\frac{{1}}{{7}}$$

Drawing two balls of same color from seven green balls can be done in $${}^7{C_2}$$ ways.
Similarly from eight white balls two can be drawn in $${}^8{C_2}$$ ways.
$$\eqalign{ & P = \frac{{{}^7{C_2}}}{{{}^{15}{C_2}}} + \frac{{{}^8{C_2}}}{{{}^{15}{C_2}}} \cr & \,\,\,\,\,\,\, = \frac{7}{{15}} \cr} $$

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that all the three marbles picked at random are blue = $$\frac{{{}^3{C_3}}}{{{}^{15}{C_3}}}$$
$$\eqalign{ & = \frac{{1 \times 3 \times 2 \times 1}}{{15 \times 14 \times 13}} \cr & = \frac{6}{{2730}} \cr & = \frac{1}{{455}} \cr} $$

Number of ways of (selecting at least two couples among five people selected) = $$\left( {{}^5{C_2} \times {}^6{C_1}} \right)$$
As remaining person can be any one among three couples left.
Required probability
$$\eqalign{ & = \frac{{{}^5{C_2} \times {}^6{C_1}}}{{{}^{10}{C_5}}} \cr & = \frac{{\left( {10 \times 6} \right)}}{{252}} \cr & = \frac{5}{{21}} \cr} $$

We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
$$\eqalign{ & = \frac{4}{{100}} \cr & = \frac{1}{{25}} \cr} $$

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
When three marbles are drawn, the probability that one is yellow and two are red
$$\eqalign{ & = \frac{{\left( {{}^2{C_1}} \right)\left( {{}^4{C_2}} \right)}}{{{}^{15}{C_3}}} \cr & = \frac{{2 \times 4 \times 3 \times 3 \times 2}}{{1 \times 2 \times 15 \times 14 \times 13}} \cr & = \frac{{144}}{{5460}} \cr & = \frac{{12}}{{455}} \cr} $$

Let E₁ be the event of drawing a red card.
Let E₂ be the event of drawing a king.
$$P\left( {{E_1} \cap {E_2}} \right) = P\left( {{E_1}} \right).P\left( {{E_2}} \right)$$
(As E₁ and E₂ are independent)
$$\eqalign{ & = \frac{1}{2} \times \frac{1}{{13}} \cr & = \frac{1}{{26}} \cr} $$

Let P(T) be the probability of getting least one tail when the coin is tossed five times.
$$P\left( {\overline T } \right)$$ = There is not even a single tail.
i.e. all the outcomes are heads.
$$\eqalign{ & P\left( {\overline T } \right) = \frac{1}{{32}} \cr & P\left( T \right) = 1 - \frac{1}{{32}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{31}}{{32}} \cr} $$

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that both marbles are blue
$$\eqalign{ & = \frac{{{}^3{C_2}}}{{{}^{15}{C_2}}} \cr & = \frac{{3 \times 2}}{{15 \times 14}} \cr & = \frac{1}{{35}} \cr} $$
Probability that both are yellow
$$\eqalign{ & = \frac{{{}^2{C_2}}}{{{}^{15}{C_2}}} \cr & = \frac{{2 \times 1}}{{15 \times 14}} \cr & = \frac{1}{{105}} \cr} $$
Probability that one blue and other is yellow
$$\eqalign{ & = \frac{{{}^3{C_1} \times {}^2{C_1}}}{{{}^{15}{C_2}}} \cr & = \frac{{2 \times 3 \times 2}}{{15 \times 14}} \cr & = \frac{2}{{35}} \cr} $$
∴ Required probability
$$\eqalign{ & \frac{1}{{35}} + \frac{1}{{105}} + \frac{2}{{35}} \cr & = \frac{{3 + 1 + 6}}{{105}} \cr & = \frac{{10}}{{105}} \cr & = \frac{2}{{21}} \cr} $$

10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!
∴ Required probability = $$\frac{{1}}{{5}}$$