Probability - Aptitude MCQ Questions and Solutions with Explanations

Here, n(S) = 52
Let E = event of getting a queen of club or a king of heart
Then, n(E) = 2
$$\eqalign{ & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{2}{{52}} \cr & = \frac{1}{{26}} \cr} $$

Let S be the sample space
Then, n(S) = number of ways of drawing 3 balls out of 15
$$\eqalign{ & = {}^{15}{C_3} \cr & = \frac{{ {15 \times 14 \times 13} }}{{ {3 \times 2 \times 1} }} \cr & = 455 \cr} $$
Let E = event of getting all the 3 red balls
$$\eqalign{ & \therefore n\left( E \right) = {}^5{C_3} = {}^5{C_2} \cr & = \frac{{ {5 \times 4} }}{{ {2 \times 1} }} = 10 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{10}}{{455}} \cr & = \frac{2}{{91}} \cr} $$

Let S be the sample space
$$\eqalign{ & n\left( S \right) = {}^{52}{C_2} = \frac{{ {52 \times 51} }}{{ {2 \times 1} }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1326 \cr} $$
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
$$\eqalign{ & {\kern 1pt} = {^{13}{C_1}{ \times ^{13}}{C_1}} \cr & {\kern 1pt} {\kern 1pt} = {13 \times 13} {\kern 1pt} {\kern 1pt} \cr & {\kern 1pt} = 169 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{169}}{{1326}} = \frac{{13}}{{102}} \cr} $$

Clearly, there are 52 cards, out of which there are 12 face cards.
$$\eqalign{ & \therefore P\left( {{\text{getting}}\,{\text{a}}\,{\text{face}}\,{\text{card}}} \right) \cr & = \frac{{12}}{{52}} = \frac{3}{{13}} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{number}}\,{\text{of}}\,{\text{balls}} \cr & = \left( {6 + 8} \right) \cr & = 14 \cr & {\text{Number}}\,{\text{of}}\,{\text{white}}\,{\text{balls}} = 8 \cr & P\left( {{\text{drawing}}\,{\text{a}}\,{\text{white ball}}} \right) \cr & = \frac{8}{{14}} = \frac{4}{7} \cr} $$

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

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 2 kings out of 4
∴ $$n(E) = {}^4\mathop C\nolimits_2 = $$   $$\frac{{\left( {4 \times 3} \right)}}{{\left( {2 \times 1} \right)}}$$   = 6
∴ $$P(E) = \frac{{n(E)}}{{n(S)}} = $$   $$\frac{6}{{1326}} = \frac{1}{{221}}$$

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 4 marbles such that none is blue.
Then, n(E) = number of ways of drawing 4 marbles out of 11 non-blue.
$${}^{11}\mathop C\nolimits_4 = \frac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2 \times 1}}$$     = 330
And n(S) = $${}^{15}\mathop C\nolimits_4 = $$   $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$     = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}}$$    $$ = \frac{{330}}{{1365}}$$  $$ = \frac{{22}}{{91}}$$
∴ Required probality =$$\left( {1 - \frac{{22}}{{91}}} \right)$$   $$ = \frac{{69}}{{91}}$$

Here S = {HH, HT, TH, TT}
Let E = event of getting at least one head = {HT, TH, HH}
$$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$

Hence, n(S) = 52
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{2}{{52}} = \frac{1}{{26}}$$