Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$\eqalign{ & {\text{In a box, }} \cr & {B_1} = 1R,\,2W\,;\,{B_2} = 2R,\,3W{\text{ and }}{B_3} = 3R,\,4W \cr & {\text{Also given that,}} \cr & P\left( {{B_1}} \right) = \frac{1}{2},\,P\left( {{B_2}} \right) = \frac{1}{3}{\text{ and }}P\left( {{B_3}} \right) = \frac{1}{6} \cr & \therefore \,P\left( {\frac{{{B_2}}}{R}} \right) \cr & = \frac{{P\left( {{B_2}} \right)P\left( {\frac{R}{{{B_2}}}} \right)}}{{P\left( {{B_1}} \right)P\left( {\frac{R}{{{B_1}}}} \right) + P\left( {{B_2}} \right)P\left( {\frac{R}{{{B_2}}}} \right) + P\left( {{B_3}} \right)P\left( {\frac{R}{{{B_3}}}} \right)}} \cr & = \frac{{\frac{1}{3} \times \frac{2}{5}}}{{\frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{5} + \frac{1}{6} \times \frac{3}{7}}} \cr & = \frac{{\frac{2}{{15}}}}{{\frac{1}{6} + \frac{2}{{15}} + \frac{1}{{14}}}} \cr & = \frac{{14}}{{39}}. \cr} $$

The probability of showing an even number in a throw $$ = \frac{3}{6} = \frac{1}{2}$$
$$\therefore $$  the required probability
$$\eqalign{ & = {}^{2n + 1}{C_1}.\frac{1}{2}.{\left( {\frac{1}{2}} \right)^{2n}} + {}^{2n + 1}{C_3}.{\left( {\frac{1}{2}} \right)^3}.{\left( {\frac{1}{2}} \right)^{2n - 2}} + ...... + {}^{2n + 1}{C_{2n + 1}}.{\left( {\frac{1}{2}} \right)^{2n + 1}} \cr & = \left( {{}^{2n + 1}{C_1} + {}^{2n + 1}{C_3} + ...... + {}^{2n + 1}{C_{2n + 1}}} \right).{\left( {\frac{1}{2}} \right)^{2n + 1}} \cr & = {2^{2n}}.{\left( {\frac{1}{2}} \right)^{2n + 1}} \cr & = \frac{1}{2} \cr} $$

The probability that the door is opened in the fifth trial
$$\eqalign{ & = \frac{9}{{10}} \times \frac{8}{9} \times \frac{7}{8} \times \frac{6}{7} \times \frac{1}{6} \cr & = \frac{1}{{10}} \cr} $$

$$\begin{array}{l}P\left(E\right)\le P\left(F\right)\Rightarrow \frac{n\left(E\right)}{n\left(S\right)}\le \frac{n\left(F\right)}{n\left(S\right)}\Rightarrow n\left(E\right)\le n\left(F\right)\\ P\left(E\cap F\right)>0\Rightarrow E\cap F\ne \varphi \\ \text{Clearly, }x\in E\cancel{\Rightarrow }x\in F,x\in F\cancel{\Rightarrow }x\in E\\ x\notin E\cancel{\Rightarrow }x\notin F\end{array}$$

$$P\left( A \right) = \frac{{3x + 1}}{3},P\left( B \right) = \frac{{1 - x}}{4}\,{\text{and }}P\left( C \right) = \frac{{1 - 2x}}{2}$$
$$\because $$ For any event $$E,0 \leqslant P\left( E \right) \leqslant 1$$
$$\eqalign{ & \Rightarrow \,\,0 \leqslant \frac{{3x + 1}}{3} \leqslant 1,0 \leqslant \frac{{1 - x}}{4} \leqslant 1\,\,{\text{and }}0 \leqslant \frac{{1 - 2x}}{2} \leqslant 1 \cr & \Rightarrow \,\, - 1 \leqslant 3x \leqslant 2, - 3 \leqslant x \leqslant 1\,\,{\text{and }} - 1 \leqslant 2x \leqslant 1 \cr & \Rightarrow \,\, - \frac{1}{3} \leqslant x \leqslant \frac{2}{3} \leqslant - 3 \leqslant x \leqslant 1,\,\,{\text{and }} - \frac{1}{2} \leqslant x \leqslant \frac{1}{2} \cr} $$
Also for mutually exclusive events $$A, B, C,$$
$$\eqalign{ & P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) \cr & \Rightarrow \,\,P\left( {A \cup B \cup C} \right) = \frac{{3x + 1}}{3} + \frac{{1 - x}}{4} + \frac{{1 - 2x}}{2} \cr & \therefore \,\,0 \leqslant \frac{{1 + 3x}}{3} + \frac{{1 - x}}{4} + \frac{{1 - 2x}}{2} \leqslant 1 \cr & 0 \leqslant 13 - 3x \leqslant 12 \cr & \Rightarrow \,\,1 \leqslant 3x \leqslant 13 \cr & \Rightarrow \,\,\frac{1}{3} \leqslant x \leqslant \frac{{13}}{3} \cr} $$
Considering all inequations, we get
$$\eqalign{ & {\text{max}}\left\{ { - \frac{1}{3}, - 3, - \frac{1}{2},\frac{1}{3}} \right\} \leqslant x \leqslant {\text{min}}\left\{ {\frac{2}{3},1,\frac{1}{2},\frac{{13}}{3}} \right\} \cr & \frac{1}{3} \leqslant x \leqslant \frac{1}{2} \cr & \Rightarrow \,\,x \in \left[ {\frac{1}{3},\frac{1}{2}} \right] \cr} $$

Let $$E :$$  ‘face $$1$$ comes up’ and
$$F :$$  ‘face $$1$$ or $$2$$ comes up’
$$\eqalign{ & \Rightarrow E \cap F = E\,\,\,\,\,\,\left( {\because \,E \subset F} \right) \cr & \therefore \,P\left( E \right) = 0.10{\text{ and}} \cr & P\left( F \right) = P\left( 1 \right) + P\left( 2 \right) \cr & = 0.10 + 0.32 \cr & = 0.42 \cr} $$
Hence, required probability
$$\eqalign{ & = P\left( {\frac{E}{F}} \right) \cr & = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}} \cr & = \frac{{P\left( E \right)}}{{P\left( F \right)}} \cr & = \frac{{0.10}}{{0.42}} \cr & = \frac{5}{{21}} \cr} $$

$$\eqalign{ & U = \left\{ {\left( {HHH} \right)\left( {HHT} \right)\left( {HTH} \right)\left( {HTT} \right)\left( {THH} \right)\left( {THT} \right)\left( {TTH} \right)\left( {TTT} \right)} \right\} \cr & A = \left\{ {\left( {TTT} \right)} \right\} \cr & B = \left\{ {\left( {HTT} \right)\left( {THT} \right)\left( {TTH} \right)} \right\} \cr & C = \left\{ {\left( {HHH} \right)\left( {HHT} \right)\left( {HTH} \right)\left( {THH} \right)} \right\} \cr & {\text{By checking the options,}}\left( {\bf{D}} \right){\text{is correct}} \cr & A \cap \left( {B' \cup C'} \right) = B' \cap C' \cr} $$

The total number of ways in which papers of $$4$$ students can be checked by seven teachers is $${7^4}.$$
The number of ways of choosing two teachers out of $$7$$ is $${}^7{C_2}.$$
The number of ways in which they can check four papers is $${2^4}.$$
But this includes two ways in which all the papers will be checked by a single teacher. Therefore, the number of ways in which $$4$$ papers can be checked by exactly two teachers is $${2^4} - 2 = 14.$$
Therefore, the number of favorable ways is $$\left( {{}^7{C_2}} \right)\left( {14} \right) = \left( {21} \right)\left( {14} \right).$$
Thus, the required probability is $$\frac{{\left( {21} \right)\left( {14} \right)}}{{{7^4}}} = \frac{6}{{49}}.$$

The probability of drawing a number less than or equal to $$15$$  in a draw $$ = \frac{{15}}{{20}} = \frac{3}{4}$$
The probability of drawing the ticket of value $$15$$  in a draw $$ = \frac{1}{{20}}$$
 $$\therefore $$  the required probability $$ = {}^4{C_1}.\frac{1}{{20}}.{\left( {\frac{3}{4}} \right)^3}$$

The probability of $$A$$ failing $$ = P\left( A \right) = \frac{1}{5}$$    and the probability of $$B$$ failing $$ = P\left( B \right) = \frac{3}{{10}}$$
The required probability
$$\eqalign{ & = P\left( {A\overline B \cup \overline A B} \right) = P\left( {A\overline B } \right) + P\left( {\overline A B} \right) \cr & = P\left( A \right).P\left( {\overline B } \right) + P\left( {\overline A } \right).P\left( B \right) \cr & = \frac{1}{5}\left( {1 - \frac{3}{{10}}} \right) + \left( {1 - \frac{1}{5}} \right).\frac{3}{{10}} \cr & = \frac{{19}}{{50}}. \cr} $$