Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$\eqalign{ & {\text{Putting}}\,n = 99{\text{ and }}p = \frac{1}{2},{\text{we have}} \cr & \left( {n + 1} \right)p = \left( {100} \right)\left( {\frac{1}{2}} \right) = 50 \cr & {\text{so that the maximum value of }}P\left( {X = r} \right) \cr & {\text{occurs at }}r = \left( {n + 1} \right) \cr & p = 50{\text{ and }}r = \left( {n + 1} \right)p - 1 = 49 \cr} $$

$$\eqalign{ & P\left( {{E_1}} \right) = \frac{{{}^6{C_3}}}{{216}} = \frac{5}{{54}}\,;\,P\left( {{E_2}} \right) = \frac{{4 + 2}}{{216}} = \frac{1}{{36}} \cr & {E_1} \cap {E_2} = {E_2} \cr & \therefore \,P\left( {{E_1} \cap {E_2}} \right) = P\left( {{E_2}} \right) = \frac{1}{{36}} \cr & P\left( {{E_2}/{E_1}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}} = \frac{{\frac{1}{{36}}}}{{\frac{5}{{54}}}} = \frac{{54}}{{180}} = \frac{3}{{10}} \cr} $$

$$n\left( S \right) = 6 \times 6 \times 6 \times 6$$
$$n\left( E \right) = $$   the number of integral solutions of $${x_1} + {x_2} + {x_3} + {x_4} = 12,$$     where $$1 \leqslant {x_1} \leqslant 6,......,1 \leqslant {x_4} \leqslant 6$$
$$\eqalign{ & = {\text{ coefficient of }}{x^{12}}{\text{ in }}{\left( {x + {x^2} + ...... + {x^6}} \right)^4} \cr & = {\text{ coefficient of }}{x^8}{\text{ in }}{\left( {\frac{{1 - {x^6}}}{{1 - x}}} \right)^4} \cr & = {\text{ coefficient of }}{x^8}{\text{ in }}{\left( {1 - {x^6}} \right)^4}.\left( {{}^3{C_0} + {}^4{C_1}x + {}^5{C_2}{x^2} + .....} \right) \cr & = {}^{11}{C_8} - 4.{}^5{C_2} = 125 \cr & \therefore \,P\left( E \right) = \frac{{125}}{{6 \times 6 \times 6 \times 6}}. \cr} $$

$${D_4}$$ can show a number appearing on one of $${D_1},$$ $${D_2}$$ and $${D_3}$$ in the following cases.
Case I : $${D_4}$$ shows a number which is shown by only one of $${D_1},$$ $${D_2}$$ and $${D_3}.$$
$${D_4}$$ shows a number in $$^6{C_1}$$ ways.
One out of $${D_1},$$ $${D_2}$$ and $${D_3}$$ can be selected in $$^3{C_1}$$ ways.
The selected die shows the same number as on $${D_4}$$ in one way and rest two dice show the different number in 5 ways each.
∴ Number of ways to happen case I
$$ = {\,^6}{C_1} \times {\,^3}{C_1} \times 1 \times 5 \times 5 = 450$$
Case II : $${D_4}$$ shows a number which is shown by only two of $${D_1},$$ $${D_2}$$ and $${D_3}.$$
As discussed in case I, it can happen in the following number of ways
$$ = {\,^6}{C_1} \times {\,^3}{C_2} \times 1 \times 1 \times 5 = 90$$
Case III : $${D_4}$$ shows a number which is shown by all three dice $${D_1},$$ $${D_2}$$ and $${D_3}.$$
Number of ways it can be done
$$ = {\,^6}{C_1} \times {\,^3}{C_3} \times 1 \times 1 \times 1 = 6$$
∴ Total number of favourable ways = 450 + 90 + 6 = 546
Also total ways $$ = 6 \times 6 \times 6 \times 6$$
∴ Required Probability $$ = \frac{{546}}{{6 \times 6 \times 6 \times 6}} = \frac{{91}}{{216}}$$

Let $$n$$ denote the required number of shots and $$X$$ the number of shots that hit the target.
Then $$X \sim B\left( {n,\,p} \right),$$    with $$p = \frac{1}{4}.$$
Now,
$$\eqalign{ & P\left( {X \geqslant 1} \right) \geqslant 0.9 \cr & \Rightarrow 1 - P\left( {X = 0} \right) \geqslant 0.9 \cr & \Rightarrow 1 - {}^n{C_0}{\left( {\frac{3}{4}} \right)^n} \geqslant 0.9 \cr & \Rightarrow {\left( {\frac{3}{4}} \right)^n} \leqslant \frac{1}{{10}} \cr & \Rightarrow {\left( {\frac{4}{3}} \right)^n} \geqslant 10 \cr & \Rightarrow n\left( {\log \,4 - \log \,3} \right) \geqslant 1 \cr & \Rightarrow n\left( {0.602 - 0.477} \right) \geqslant 1 \cr & \Rightarrow n \geqslant \frac{1}{{0.125}} = 8 \cr} $$
Therefore the least number of trials required is $$8.$$

$$\eqalign{ & P\left( {\frac{{\overline A }}{{\overline B }}} \right) = \frac{{P\left( {\overline A \cap \overline B } \right)}}{{P\left( {\overline B } \right)}} \cr & = \frac{{P\left( {\overline {A \cup B} } \right)}}{{P\left( {\overline B } \right)}} \cr & = \frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}} \cr} $$

The required probability
$$\eqalign{ & = P\left( {A\overline B \cup \overline A B} \right) \cr & = P\left( A \right)P\left( {\overline B } \right) + P\left( {\overline A } \right)P\left( B \right) \cr & = \frac{3}{{10}}\left( {1 - \frac{2}{5}} \right) + \left( {1 - \frac{3}{{10}}} \right)\frac{2}{5} \cr & = \frac{{23}}{{50}} \cr} $$

$$\eqalign{ & \because \,P\left( {\overline B } \right) = 0.8 \Rightarrow P\left( B \right) = 0.2 \cr & P\left( {A \cup B} \right) = 0.5\,\,\& \,\,P\left( {A|B} \right) = 0.4 \cr & \because \,P\left( {A \cap B} \right) = P\left( B \right) \times P\left( {A|B} \right) \cr & \Rightarrow P\left( {A \cap B} \right) = 0.2 \times 0.4 \cr & \Rightarrow P\left( {A \cap B} \right) = 0.08 \cr & \& \,\,P\left( A \right) = P\left( {A \cup B} \right) - P\left( B \right) + P\left( {A \cap B} \right) \cr & \Rightarrow P\left( A \right) = 0.5 - 0.2 + 0.08 \cr & \Rightarrow P\left( A \right) = 0.38 \cr} $$

$$\eqalign{ & {\text{Probability of }}A = \frac{1}{6} \cr & {\text{Probability of }}B = \frac{1}{5} \cr & {\text{Probability of }}C = \frac{1}{4} \cr & {\text{Hence,}} \cr & {\text{Required probability}} = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} = \frac{1}{{120}} \cr} $$

Total number of ways of selecting $$3$$ integers from $$20$$  natural numbers $$ = {}^{20}{C_3} = 1140$$
Their product is a multiple of $$3$$ means, at least one number is divisible by $$3.$$
The numbers which are divisible by $$3$$ are $$3,\,6,\,9,\,12,\,15,\,18$$     and the number of ways of selecting at least one of them is
$${}^6{C_1} \times {}^{14}{C_2} + {}^6{C_2} \times {}^{14}{C_1} + {}^6{C_3} = 776$$
Required Probability $$ = \frac{{776}}{{1140}} = \frac{{194}}{{285}}$$