Probability MCQ Questions & Answers in Statistics and Probability | Maths

The total number of ways of choosing $$11$$ players out of $$15$$ is $${}^{15}{C_{11}}.$$  A team of $$11$$ players containing at least $$3$$ bowlers can be chosen in the following mutually exclusive ways :
$$\left( {{\bf{I}}} \right)$$  Three bowlers out of $$5$$ bowlers and $$8$$ other players out of the remaining $$10$$ players.
$$\left( {{\bf{II}}} \right)$$  Four bowlers out of $$5$$ bowlers and $$7$$ other players out of the remaining $$10$$ players.
$$\left( {{\bf{III}}} \right)$$  Five bowlers out of $$5$$ bowlers and $$6$$ other players out of the remaining $$10$$ players.
So, required probability
$$\eqalign{ & = P\left( {\bf{I}} \right) + P\left( {{\bf{II}}} \right) + P\left( {{\bf{III}}} \right) \cr & = \frac{{{}^5{C_3} \times {}^{10}{C_8}}}{{{}^{15}{C_{11}}}} + \frac{{{}^5{C_4} \times {}^{10}{C_7}}}{{{}^{15}{C_{11}}}} + \frac{{{}^5{C_5} \times {}^{10}{C_6}}}{{{}^{15}{C_{11}}}} \cr & = \frac{{12}}{{13}} \cr} $$

$$\eqalign{ & P\left( Y \right) = \frac{3}{6} = \frac{1}{2},\,\,\,\,\,P\left( R \right) = \frac{2}{6} = \frac{1}{3},\,\,\,\,\,P\left( B \right) = \frac{1}{6} \cr & \therefore \,P\left( {Y \cap R \cap B} \right) = P\left( Y \right).P\left( R \right).P\left( B \right) \cr & = \frac{1}{2}.\frac{2}{6}.\frac{1}{6} \cr & = \frac{1}{{36}} \cr} $$

The first object can be given to any of the $$n$$ persons. But the second, third and other objects, too, can go to any of the $$n$$ persons. Therefore the total number of ways of distributing the $$n$$ objects randomly among $$n$$ persons is $${n^n}.$$
There are $${}^n{P_n} = n!$$   ways in which each person gets exactly one object, so the probability of this happening is $$\frac{{n!}}{{{n^n}}} = \frac{{\left( {n - 1} \right)!}}{{{n^{n - 1}}}}.$$
Hence the probability that at least one person does not get any object is $$1 - \frac{{\left( {n - 1} \right)!}}{{{n^{n - 1}}}}.$$

The probability of $$A$$ winning in a game $$ = P\left( A \right) = \frac{6}{{12}} = \frac{1}{2}$$
The probability of $$B$$ winning in a game $$ = P\left( B \right) = \frac{4}{{12}} = \frac{1}{3}$$
The required probability
$$\eqalign{ & = P\left( {A \cap B \cap A} \right) + P\left( {B \cap A \cap B} \right) \cr & = P\left( A \right).P\left( B \right).P\left( A \right) + P\left( B \right).P\left( A \right).P\left( B \right) \cr & = \frac{1}{2}.\frac{1}{3}.\frac{1}{2} + \frac{1}{3}.\frac{1}{2}.\frac{1}{3} \cr & = \frac{5}{{36}}. \cr} $$ 

Let $$G \equiv $$  original signal is green
⇒ $$P(G) = \frac{4}{5}$$
$${E_1}$$ $$ \equiv $$ $$A$$ receives the signal correctly $$P\left( {{E_1}} \right) = \frac{3}{4}$$
$${E_2}$$ $$ \equiv $$ $$B$$ receives the signal correctly $$P\left( {{E_2}} \right) = \frac{3}{4}$$
$$E \equiv $$  Signal received by $$B$$ is green.
Then $$E$$ can happen in the following ways

Original Signal	Received at A	Received at B
Red $$ \to $$	Red $$ \to $$	Green
Red $$ \to $$	Green $$ \to $$	Green
Green $$ \to $$	Green $$ \to $$	Green
Green $$ \to $$	Red $$ \to $$	Green

$$\eqalign{ & \therefore \,\,P\left( E \right) = \left( {\overline G \cap {E_1} \cap \overline {{E_2}} } \right) + \left( {\overline G \cap {E_1} \cap \overline {{E_2}} } \right) + P\left( {G \cap {E_1} \cap {E_2}} \right) + P\left( {G \cap \overline {{E_1}} \cap \overline {{E_2}} } \right) \cr & = \frac{1}{5} \times \frac{3}{4} \times \frac{1}{4} + \frac{1}{5} \times \frac{3}{4} \times \frac{1}{4} + \frac{4}{5} \times \frac{3}{4} \times \frac{3}{4} + \frac{4}{5} \times \frac{1}{4} \times \frac{1}{4} \cr & = \frac{{3 + 3 + 36 + 4}}{{80}} \cr & = \frac{{46}}{{80}} \cr & = \frac{{23}}{{40}} \cr} $$
$$\eqalign{ & \therefore \,\,P\left( {\frac{G}{E}} \right) = \frac{{P\left( {G \cap E} \right)}}{{P\left( E \right)}} \cr & = \frac{{P\left( {G \cap {E_1} \cap {E_2}} \right) + P\left( {G \cap {{\overline E }_1} \cap {{\overline E }_2}} \right)}}{{P\left( E \right)}} \cr & = \frac{{\frac{4}{5} \times \frac{3}{4} \times \frac{3}{4} + \frac{4}{5} \times \frac{1}{4} \times \frac{1}{4}}}{{\frac{{23}}{{40}}}} \cr & = \frac{{\frac{{40}}{{80}}}}{{\frac{{23}}{{40}}}} \cr & = \frac{{20}}{{23}} \cr} $$

$$\eqalign{ & \,\,\,\,\,P\left( {A \cup B \cup C} \right) \cr & = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) - P\left( {A \cap B \cap C} \right) \cr & = 0.6 + 0.4 + 0.5 - 0.2 - P\left( {B \cap C} \right) - 0.3 + 0.2 \cr & = 1.2 - P\left( {B \cap C} \right) \cr} $$
$$\eqalign{ & {\text{because }}P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr & \Rightarrow 0.8 = 0.6 + 0.4 - P\left( {A \cap B} \right) \cr & {\text{But }}0.85 \leqslant P\left( {A \cup B \cup C} \right) \leqslant 1 \cr & \therefore \,0.85 \leqslant 1.2 - P\left( {B \cap C} \right) \leqslant 1 \cr & \Rightarrow 0.2 \leqslant P\left( {B \cap C} \right) \leqslant 0.35 \cr} $$

Prob. of a getting a white ball in a single draw
$$ = p = \frac{{12}}{{24}} = \frac{1}{2}$$
Prob. of getting a white ball $${4^{th}}$$ time in the $${7^{th}}$$ draw
= $$P$$ (getting 3 $$W$$ in 6 draws) $$ \times $$ $$P$$ (getting $$W$$ ball at $${7^{th}}$$ draw)
$$ = {\,^6}{C_3}{\left( {\frac{1}{2}} \right)^6}.\frac{1}{2} = \frac{5}{{32}}$$

$$\eqalign{ & n\left( S \right) = {}^7{C_3} \times 3! = \frac{{7.6.5}}{6}.6 = 210 \cr & n\left( E \right) = {}^2{C_1} \times {}^2{C_1} \times {}^1{C_1} \times 3!, \cr} $$
because one has to sit at any one of the two marked seats on the left and the other has to sit at any one of the two marked seats on the right.
$$\therefore \,P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{2 \times 2 \times 6}}{{210}} = \frac{4}{{35}}.$$

$$\eqalign{ & {\text{We have, }}P\left( {x \geqslant 1} \right) \geqslant \frac{9}{{10}} \cr & \Rightarrow 1 - P\left( {x = 0} \right) \geqslant \frac{9}{{10}} \cr & \Rightarrow 1 - {}^n{C_0}{\left( {\frac{1}{4}} \right)^0}{\left( {\frac{3}{4}} \right)^n} \geqslant \frac{9}{{10}} \cr & \Rightarrow 1 - \frac{9}{{10}} \geqslant {\left( {\frac{3}{4}} \right)^n} \cr & \Rightarrow {\left( {\frac{3}{4}} \right)^n} \leqslant \left( {\frac{1}{{10}}} \right) \cr & {\text{Taking log to the base }}\frac{3}{4}{\text{, on both sides, we get}} \cr & n\,{\log _{\frac{3}{4}}}\left( {\frac{3}{4}} \right) \geqslant {\log _{\frac{3}{4}}}\left( {\frac{1}{{10}}} \right) \cr & \Rightarrow n \geqslant - {\log _{\frac{3}{4}}}10 = \frac{{ - {{\log }_{10}}10}}{{{{\log }_{10}}\left( {\frac{3}{4}} \right)}} \cr & \Rightarrow n \geqslant \frac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}} \cr} $$

Let us divide the players into two pools $$A$$ and $$B$$ each containing $$4$$ players.
Let $${P_4}$$ be in pool $$A$$. Now $${P_4}$$ will reach the final if we fill the remaining three of pool $$A$$ by any of $${P_5},\,{P_6},\,{P_7}{\text{ or }}{P_8}$$
$$\therefore $$  Probability is $$\frac{{{}^4{C_3}}}{{{}^7{C_3}}} = \frac{{4.3.2}}{{7.6.5}} = \frac{4}{{35}}$$