Probability - Aptitude MCQ Questions and Solutions with Explanations

Number of Red balls = 4
Number of Blue balls = 6
Number of Pink balls = 8
Total number of balls = 4 + 6 + 8 = 18
Required probability
$$\eqalign{ & = \frac{4}{{18}} \times \frac{{11}}{{20}} + \frac{6}{{18}} \times \frac{{11}}{{20}} \cr & = \frac{{11}}{{20}}\left[ {\frac{4}{{18}} + \frac{6}{{18}}} \right] \cr & = \frac{{11}}{{20}} \times \frac{{10}}{{18}} \cr & = \frac{{11}}{{36}} \cr} $$

Number of blue balls = 3
Number of green balls = 2
Numbers of red balls = 5
Total balls in the bag = 3 + 2 + 5 = 10
Total possible outcomes = Selection of 4 balls out of 10 balls
$$ = {}^{10}\mathop C\nolimits_4 = \frac{{10!}}{{4! \times (10 - 4)!}}$$
$$ = \frac{{10 \times 9 \times 8 \times 7}}{{1 \times 2 \times 3 \times 4}}$$
= 210
Favorable outcomes = (selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls)
$$\eqalign{ & = {}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_2 \cr & = 1 \times 3 \cr & = 3 \cr} $$
∴ Required probability
  $$ = \frac{{{\text{Favorable outcomes}}}}{{{\text{Total possible outcomes}}}}$$
$$\eqalign{ & = \frac{3}{{210}} \cr & = \frac{1}{{70}} \cr} $$

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 5 balls out of 9 non-blue balls.
$$ = {}^9\mathop C\nolimits_5 = {}^9\mathop C\nolimits_{\left( {9 - 5} \right)} = {}^9\mathop C\nolimits_4 $$    $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$    = 126
And,
n(S) = $${}^{15}\mathop C\nolimits_5 = $$   $$\frac{{15 \times 14 \times 13 \times 12 \times 11}}{{5 \times 4 \times 3 \times 2 \times 1}}$$     = 3003
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = $$    $$\frac{{126}}{{3003}} = \frac{6}{{143}}$$
∴ Required Probability
$$\eqalign{ & = \left( {1 - \frac{6}{{143}}} \right) \cr & = \frac{{137}}{{143}} \cr} $$

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{8} = \frac{1}{2}$$

In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36
Let E = event of getting a total of 10 or 11
= [(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)]
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{5}{{36}}$$

Hence, n (S) = 52
There are 13 spades (including one ten) and there are 3 more ten
Let E = event of getting a ten or a spade
Then, n (E) = (13 + 3) = 16
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{16}}{{52}} = \frac{4}{{13}}$$

Total number of balls = (4 + 5 + 6) = 15
P (drawing a red ball or a green ball) = P (red) + P (green)
$$\eqalign{ & = \left( {\frac{4}{{15}} + \frac{5}{{15}}} \right) \cr & = \frac{9}{{15}} \cr & = \frac{3}{5} \cr} $$

Here S = [TTT, TTH, THT, HTT, THH, HTH, HHT, HHH]
Let E = event of getting at most two heads
Then, E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{8}$$

Number of red balls = 6
Number of yellow balls = 11
Number of pink balls = 5
Total number of balls = 6 + 11 + 5 = 22
Total possible outcomes
$$n(E) = {}^{22}\mathop C\nolimits_2 = \frac{{22!}}{{2!(22 - 2)!}}$$
$$ = \frac{{22!}}{{2! \times 20!}} $$
$$ = \frac{{22 \times 21}}{{2 \times 1}} $$
$$=$$ 231
Number of favourable outcomes
$$n(S) = {}^6\mathop C\nolimits_1 \times {}^{11}\mathop C\nolimits_1 $$    = 6 × 11 = 66
Required probability = $$\frac{{n(E)}}{{n(S)}}$$   $$ = \frac{{66}}{{231}}$$   $$ = \frac{1}{7}$$

Let $${E_1}$$ = event that the husband is selected and $${E_2}$$ = event that the wife is selected.
Then, $$P({E_1}) = \frac{1}{7}{\text{ }}$$   and $${\text{ }}P({E_2}) = \frac{1}{5}$$
∴ $$P({\overline E _1}) = \left( {1 - \frac{1}{7}} \right)$$   $${\text{ = }}\frac{6}{7}$$
and
$$P({\overline E _2}) = $$   $$\left( {1 - \frac{1}{5}} \right) = \frac{4}{5}$$
∴ Required probability = P [( A and not B) or (B and not A)]
= P $$\left[ {\left( {{E_1} \cap {{\bar E}_2}} \right)\,{\text{or}}\,\left( {{E_2} \cap {{\bar E}_1}} \right)} \right]$$
= P $$\left( {{E_1} \cap {{\overline E }_2}} \right) + {\text{ P}}\left( {{E_2} \cap {{\overline E }_1}} \right)$$
= $${\text{P}}\left( {{E_1}} \right).{\text{ P}}\left( {{{\overline E }_2}} \right) + P\left( {{E_2}} \right).P\left( {{{\overline E }_1}} \right)$$
$$\eqalign{ & = \left( {\frac{1}{7} \times \frac{4}{5}} \right) + \left( {\frac{1}{5} \times \frac{6}{7}} \right) \cr & = \frac{10}{{35}} \cr & = \frac{2}{7} \cr} $$