Probability - Aptitude MCQ Questions and Solutions with Explanations

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 2 blue and 1 yellow marble.
Then, n(E) = $$\left( {{}^4\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right)$$   $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times 3} \right)$$   = 18
Also, n(S) = $${}^{15}\mathop C\nolimits_3 = $$   $$\frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$   = 455
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{18}}{{455}}$$

n(S) = number of ways of choosing 4 persons out of 9
$$ = {}^9\mathop C\nolimits_4 $$   $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$   = 126
n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E) $$ = \left( {{}^4\mathop C\nolimits_2 \times {}^5\mathop C\nolimits_2 } \right)$$   $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$     = 60
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 3 non-yellow balls
Then, n(E) = $${}^{12}\mathop C\nolimits_3 $$   $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$   = 220
Also, n(S) = $${}^{15}\mathop C\nolimits_3 $$   $$ = \frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$   = 455
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{220}}{{455}} = \frac{{44}}{{91}}$$

Here, n(S) = 52
There are 13 cards of diamond (including one king) and there are three more kings.
Let E = event of getting a diamond or a king
Then, n(E) = (13 + 3) = 16
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{16}}{{52}} = \frac{4}{{13}}$$

In two throws of dice, n(S) = (6 × 6) = 36
Let E = event of getting a sum 9 = [(3, 6), (4, 5), (5, 4), (6, 3)]
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{36}} = \frac{1}{9}$$

Let S be the sample space.
Then, n(S) = $${}^{52}\mathop C\nolimits_2 $$  $$ = \frac{{\left( {52 \times 51} \right)}}{{\left( {2 \times 1} \right)}}$$   = 1326
Let E = event of getting 1 spade and 1 heart.
∴ n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13
$$ = \left( {{}^{13}\mathop C\nolimits_1 \times {}^{13}\mathop C\nolimits_1 } \right)$$   = (13 × 13) = 169
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{169}}{{1326}} = \frac{{13}}{{102}}$$

Total number of persons = (3 + 2) = 5
$$\therefore n(S) = {}^5\mathop C\nolimits_3 = {}^5\mathop C\nolimits_2 $$     $$ = \frac{{5 \times 4}}{{2 \times 1}}$$   = 10
Let E be the event of selecting 3 members having at least 1 women
Then, n(E) = n [(1 women and 2 men ) or (2 women and 1 man)]
= n (1 woman and 2 men) + n (2 women and 1 man)
$$\eqalign{ & = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_2 } \right) + \left( {{}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right) \cr & = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_1 } \right) + \left( {1 \times {}^3\mathop C\nolimits_1 } \right) \cr & = \left( {2 \times 3} \right) + \left( {1 \times 3} \right) \cr & = \left( {6 + 3} \right) \cr & = 9 \cr} $$
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{10}}$$

Let $${{{\text{E}}_1}}$$ = event that A speaks the truth
And $${{{\text{E}}_2}}$$ = event that B speaks the truth
Then,
$$\eqalign{ & P\left( {{E_1}} \right) = \frac{{75}}{{100}} = \frac{3}{4}, \cr & P\left( {{E_2}} \right) = \frac{{80}}{{100}} = \frac{4}{5}, \cr & P\left( {{{\overline E }_1}} \right) = \left( {1 - \frac{3}{4}} \right) = \frac{1}{4}, \cr & P\left( {{{\overline E }_2}} \right) = \left( {1 - \frac{4}{5}} \right) = \frac{1}{5} \cr} $$
P (A and B contradict each other)
= P [(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)]
$$\eqalign{ & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right)or\left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right) + \left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right) + P\left( {{{\overline E }_1}} \right).P\left( {{E_2}} \right) \cr & = \left( {\frac{3}{4} \times \frac{1}{5}} \right) + \left( {\frac{1}{4} \times \frac{4}{5}} \right) \cr & = \left( {\frac{3}{{20}} + \frac{1}{5}} \right) \cr & = \frac{7}{{20}} \cr & = \left( {\frac{7}{{20}} \times 100} \right)\% \cr & = 35\% \cr} $$

Total number of balls = (4 + 5 + 3) = 12
Let E be the event of drawing 2 red balls.
Then, n (E) = $${}^4\mathop C\nolimits_2 = \frac{{4 \times 3}}{{2 \times 1}}$$   = 6
Also n (S) = $${}^{12}\mathop C\nolimits_2 = \frac{{12 \times 11}}{{2 \times 1}}$$   = 66
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{66}} = \frac{1}{{11}}$$

P (getting a prize)
$$\eqalign{ & = \frac{{10}}{{\left( {10 + 25} \right)}} \cr & = \frac{{10}}{{35}} \cr & = \frac{2}{7} \cr} $$