Probability - Aptitude MCQ Questions and Solutions with Explanations

In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36
Let E = event of getting two numbers whose product is even.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n (E) = 27
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{27}}{{36}} = \frac{3}{4}$$

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 1 green, 2 blue and 1 red marble.
Then,
n (E) = $${}^2\mathop C\nolimits_1 \times {}^4\mathop C\nolimits_2 \times {}^6\mathop C\nolimits_1 $$   $$ = 2 \times \frac{{4 \times 3}}{{2 \times 1}} \times 6$$     = 72
And, n (S) = $${}^{15}\mathop C\nolimits_4 = $$   $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$     = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{72}}{{1365}}$$      $$ = \frac{{24}}{{455}}$$

Total number of balls = (2 + 3 + 2) = 7
Let, E be the event of drawing 2 non-blue balls.
Then, n (E) = $${}^5\mathop C\nolimits_4 = \frac{{5 \times 4}}{{2 \times 1}}$$   = 10
And, n (S) = $${}^7\mathop C\nolimits_2 = \frac{{7 \times 6}}{{2 \times 1}}$$   = 21
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{10}}{{21}}$$

Clearly, n (S) = (6 × 6) = 36
Let E be the event that the sum is a prime number. Then, n (E) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3,), (5, 2), (5, 6), (6, 1), (6, 5)}
∴ n (E) = 15
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{36}} = \frac{5}{{12}}$$

Total number of balls = (10 + 10) = 20
Let E be the event of drawing 2 balls of the same colour.
n (E) = number of ways of drawing 2 black balls or 2 white balls.
n (E) =   $$\left( {{}^{10}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_2 } \right)$$   $$ = 2 \times {}^{10}\mathop C\nolimits_2 $$   $$ = 2 \times \frac{{10 \times 9}}{{2 \times 1}}$$   = 90
n (S) = number of ways of drawing 2 balls out of 20 balls
$$ = {}^{20}\mathop C\nolimits_2 = \frac{{20 \times 19}}{{2 \times 1}}$$   = 190
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{90}}{{190}} = \frac{9}{{19}}$$

Out of mangoes, 4 mangoes are rotten
∴ Required probability
$$\eqalign{ & = \frac{{{}^6{C_2}}}{{{}^{10}{C_2}}} \cr & = \frac{{\frac{{6!}}{{2!\left( {6 - 2} \right)!}}}}{{\frac{{10!}}{{2!\left( {10 - 2} \right)!}}}} \cr & = \frac{{\frac{{6!}}{{2!4!}}}}{{\frac{{10!}}{{2! \times 8!}}}} \cr & = \frac{{\frac{{6 \times 5}}{{1 \times 2}}}}{{\frac{{10 \times 9}}{{1 \times 2}}}} \cr & = \frac{{6 \times 5}}{{10 \times 9}} \cr & = \frac{1}{3} \cr} $$

Let $${E_1}$$ = event that A speaks the truth
And $${E_2}$$ = event that B speaks the truth
Then,
$$\eqalign{ & P\left( {{E_1}} \right) = \frac{{60}}{{100}} = \frac{3}{5}, \cr & P\left( {{E_2}} \right) = \frac{{70}}{{100}} = \frac{7}{{10}}, \cr & P\left( {{{\bar E}_1}} \right) = \left( {1 - \frac{3}{5}} \right) = \frac{2}{5}, \cr & P\left( {{{\bar E}_2}} \right) = \left( {1 - \frac{7}{{10}}} \right) = \frac{3}{{10}} \cr} $$
P (A and B say the same thing) = P [(A speaks the truth and B speaks the truth) or (A tells a lie and B tells a lie)]
$$=$$ P [$$\left( {{E_1} \cap {E_2}} \right)$$   or   $$({\overline E _1} \cap {\overline E _2})] $$
$$=$$ P $$\left( {{E_1} \cap {E_2}} \right)$$   + $$({\overline E _1} \cap {\overline E _2})$$
$$=$$ P ($${{E_1}}$$). P ($${{E_2}}$$) + P ($${\overline E _1}$$) . P ($${\overline E _2}$$)
$$\eqalign{ & = \left( {\frac{3}{5} \times \frac{7}{{10}}} \right) + \left( {\frac{2}{5} \times \frac{3}{{10}}} \right) \cr & = \frac{{27}}{{50}} \cr & = 0.54 \cr} $$

Let S be the sample space and let E be the event of selecting 2 boys and 1 girl.
Then, n(S) = number of ways of selecting 3 students out of 25
= $${}^{25}\mathop C\nolimits_3 = $$   $$\frac{{25 \times 24 \times 23}}{{3 \times 2 \times 1}}$$   = 2300
And, n(E) = $$\left( {{}^{15}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_1 } \right)$$   $$ = \left( {\frac{{15 \times 14}}{{2 \times 1}} \times 10} \right)$$    = 1050
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}$$

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 4 balls such that 2 are red and 2 are green.
Then, n(E) = $$\left( {{}^2\mathop C\nolimits_2 \times {}^4\mathop C\nolimits_2 } \right)$$   $$ = \left( {1 \times \frac{{4 \times 3}}{{2 \times 1}}} \right)$$   = 6
And, n(S) = $${}^{15}\mathop C\nolimits_4 = $$   $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$     = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{1365}} = \frac{2}{{455}}$$

$$\eqalign{ & P(E) = \frac{{30}}{{100}} = \frac{3}{{10}}; \cr & P(H) = \frac{{20}}{{100}} = \frac{1}{5}\,\,{\text{and}} \cr & {\text{P }}\left( {{\text{E}} \cap {\text{H}}} \right)\,{\text{ = }}\frac{{10}}{{100}} = \frac{1}{{10}} \cr} $$
P (E or H) = $$P(E \cup H)$$
= P(E) + P(H) - $$P(E \cup H)$$
$$\eqalign{ & = \left( {\frac{3}{{10}} + \frac{1}{5} - \frac{1}{{10}}} \right) \cr & = \frac{4}{{10}} \cr & = \frac{2}{5} \cr} $$