Probability - Aptitude MCQ Questions and Solutions with Explanations

We know that in a simultaneous throw of two dice,
n(S) = 6 × 6 = 36
Let E = event of getting a total of 7
= {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{36}} = \frac{1}{6}$$

Here, n(S) = 52
There are 26 red cards (including 2 kings) and there are 2 more kings.
Let E = event of getting a red card or a king.
Then, n(E) = 28
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{28}}{{52}} = \frac{7}{{13}}$$

Clearly,
n (S) = $$n{\text{ }}(S) = $$   $${}^{52}\mathop C\nolimits_2 = $$   $$\frac{{\left( {52 \times 51} \right)}}{2}$$   = 1326
Let $${{E_1}}$$ = event of getting both red cards
$${{E_2}}$$ = event of getting both kings
Then, $${{E_1}}$$ $$ \cap $$ $${{E_2}}$$ = event of getting 2 kings of red cards.
∴ $$n{\text{ }}({E_1}) = {}^{26}\mathop C\nolimits_2 = \frac{{\left( {26 \times 25} \right)}}{{\left( {2 \times 1} \right)}}$$     = 325 and
$$n{\text{ }}({E_2}) = {}^4\mathop C\nolimits_2 = \frac{{\left( {4 \times 3} \right)}}{{\left( {2 \times 1} \right)}}$$     = 6
$$n\left( {{E_1} \cap {E_2}} \right) = {}^2{C_2} = 1$$
$$\therefore P({E_1}) = \frac{{n({E_1})}}{{n(S)}} = \frac{{325}}{{1326}}$$      and
$$P({E_2}) = \frac{{n({E_2})}}{{n(S)}} = \frac{6}{{1326}}$$
$$P({E_1} \cap {E_2}) = \frac{1}{{1326}}$$
∴ P (both red or both kings)
$$\eqalign{ & = P\left( {{E_1} \cup {E_2}} \right) \cr & = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) - P\left( {{E_1} \cap {E_2}} \right) \cr & = \left( {\frac{{325}}{{1326}} + \frac{6}{{1326}} - \frac{1}{{1326}}} \right) \cr & = \frac{{330}}{{1326}} \cr & = \frac{{55}}{{221}} \cr} $$

Total number of balls = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 2 red balls.
Then, n(E)  $$ = {}^6\mathop C\nolimits_2 $$  $$ = \frac{{6 \times 5}}{{2 \times 1}}$$   = 15
Also, $$n(S) = {}^{15}\mathop C\nolimits_2 $$   $$ = \frac{{15 \times 14}}{{2 \times 1}}$$   = 105
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{105}} = \frac{1}{7}$$

Total number of marbles = (4 + 5 + 3) = 12
Let E be the event of drawing 3 marbles such that none is blue.
Then, n (E) = number of ways of drawing 3 marbles out of 7 = $${}^7\mathop C\nolimits_3 $$   $$ = \frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}}$$   = 35
And, $$n(S) = {}^{12}\mathop C\nolimits_3 $$   $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$   = 220
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{35}}{{220}} = \frac{7}{{44}}$$
∴ Required probability
= 1 - P(E)
= $$\left( {1 - \frac{7}{{44}}} \right)$$
= $$ \frac{{37}}{{44}}$$

P (none is defective)
= n (E) =  $$\frac{{{}^{16}\mathop C\nolimits_2 }}{{{}^{20}\mathop C\nolimits_2 }} = $$ $$\left( {\frac{{16 \times 15}}{{2 \times 1}} \times \frac{{2 \times 1}}{{20 \times 19}}} \right)$$     $$ = \frac{{12}}{{19}}$$
P (at least 1 is defective)
=$$\left( {1 - \frac{{12}}{{19}}} \right)$$   $$ = \frac{7}{{19}}$$

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 2 marbles such that either both are green or both are yellow.
Then,
n (E) = $$\left( {{}^2\mathop C\nolimits_1 + {}^3\mathop C\nolimits_2 } \right)$$   $$ = \left( {1 + {}^3\mathop C\nolimits_1 } \right)$$   = (1 + 3) = 4
And, n (S) = $${}^{15}\mathop C\nolimits_2 = $$   $$\frac{{15 \times 14}}{{2 \times 1}}$$   = 105
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$

Here, S = {1, 2, 3, 4, ....., 19, 20}
Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{20}}$$

When a die is thrown, we have S = {1, 2, 3, 4, 5, 6}
Let, E = event of getting a number greater than 4 = {5, 6}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{2}{6} = \frac{1}{3}$$

Clearly, there are 52 cards, out of which there are 12 face cards 4 jack, 4 queens, and 4 kings
∴ P (getting a face card) $$ = \frac{{12}}{{52}}$$   $$ = \frac{3}{{13}}$$