Probability - Aptitude MCQ Questions and Solutions with Explanations

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that at least one green marble can be picked in the random draw of two marbles = Probability that one is green + Probability that both are green
$$\eqalign{ & = \frac{{{}^6{C_1}\, \times \,{}^9{C_1}}}{{{}^{15}{C_2}}} + \frac{{{}^6{C_2}}}{{{}^{15}{C_2}}} \cr & = \frac{{6 \times 9 \times 2}}{{15 \times 14}} + \frac{{6 \times 5}}{{15 \times 14}} \cr & = \frac{{36}}{{70}} + \frac{1}{7} \cr & = \frac{{46}}{{70}} \cr & = \frac{{23}}{{35}} \cr} $$

Since four coins are tossed, sample space = 2⁴
Getting two heads and two tails can happen in six ways.
n(E) = six ways
p(E) = $$\frac{6}{{{2^4}}}$$ = $$\frac{3}{8}$$

Possible event = 6 × 6 = 36
Favourable outcomes = {(3, 6), {4, 5}, {5, 4}, (6, 3)}
∴ Probability
$$\eqalign{ & = \frac{{{\text{Favourable}}}}{{{\text{total}}}} \cr & = \frac{4}{{36}} \cr & = \frac{1}{9} \cr} $$

Here, one probability is to find which pot is selected and other is for red marble from the selected pot,
Probability of selecting 1 post out of 2 pots = $$\frac{{1}}{{2}}$$
Say it has 4 red and 2 green marbles
So, Red marble probability = $$\frac{{4}}{{4 + 2}}$$ = $$\frac{{4}}{{6}}$$
So Probability 1 $$ = \frac{1}{2} \times \frac{4}{6} = \frac{4}{{12}}$$
Similarly, Probability 2 $$ = \frac{1}{2} \times \frac{5}{{5 + 3}} = \frac{5}{{16}}$$
Total probability of red ball
$$\eqalign{ & = \frac{4}{{12}} + \frac{5}{{16}} \cr & = \frac{{31}}{{48}} \cr} $$

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
E = {HTT, THT, TTH, TTT}
$$\eqalign{ & {\text{n(S) = 8}} \cr & {\text{n(E) = 4}} \cr & {\text{P(E) = }}\frac{{{\text{n(E)}}}}{{{\text{n(S)}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\text{4}}}{8} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{0}}{\text{.5}} \cr} $$

Total cards = 52 cards
NUmber of club cards = 13 Probability of 1^st card being club card = $$\frac{{13}}{{52}}$$
Probability of 2^nd card being club card = $$\frac{{12}}{{51}}$$
∴ Total Probability
$$\eqalign{ & = \frac{{13}}{{52}} \times \frac{{12}}{{51}} \cr & = \frac{1}{{17}} \cr} $$

A non-leap year has 365 days, which has 52 weeks (364 days) means 52 Thursdays
Thus there is just 1 day extra
We want it to be Thursday
Total possibilities are 7 (Sunday to Saturday means 7 days)
∴ Probability of 53 Thursday = $$\frac{{1}}{{7}}$$

We want 2 black shirts
That means choose one and then choose other from remaining shirts
There are 6 black shirts
Total 15 shirts
So probability for 2 black shirts
$$\eqalign{ & = \frac{6}{{15}} \times \frac{5}{{14}} \cr & = \frac{1}{7} \cr} $$

We want jack or spade
There are 4 jacks and 13 spade cards
Total 52 cards
Also 1 jack is jack of spades
So probability
$$\eqalign{ & = \frac{4}{{52}} + \frac{{13}}{{52}} - \frac{1}{{52}} \cr & = \frac{4}{{13}} \cr} $$

S = {1, 2, 3, ........, 49, 50}
E = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 7, 14, 21, 35, 42, 49}
$$\eqalign{ & n\left( S \right) = 50 \cr & n\left( E \right) = 18 \cr & P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{50}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{9}{{25}} \cr} $$