Probability MCQ Questions & Answers in Statistics and Probability | Maths

\[\left. \begin{array}{l} np = 4\\ npq = 2 \end{array} \right\}\]
$$\eqalign{ & \Rightarrow \,\,q = \frac{1}{2},p = \frac{1}{2},n = 8 \cr & P\left( {X = 1} \right) = \,{\,^8}{C_1}\left( {\frac{1}{2}} \right){\left( {\frac{1}{2}} \right)^7} \cr & = 8.\frac{1}{{{2^8}}} \cr & = \frac{1}{{{2^5}}} \cr & = \frac{1}{{32}} \cr} $$

$$\eqalign{ & p\left( {X \geqslant 8} \right) \cr & = {}^{10}{C_8}{\left( {\frac{1}{2}} \right)^8}{\left( {\frac{1}{2}} \right)^2} + {}^{10}{C_9}{\left( {\frac{1}{2}} \right)^9}{\left( {\frac{1}{2}} \right)^1} + {}^{10}{C_{10}}{\left( {\frac{1}{2}} \right)^{10}} \cr & = {\left( {\frac{1}{2}} \right)^{10}}\left[ {{}^{10}{C_2} + {}^{10}{C_1} + {}^{10}{C_0}} \right] \cr & = \frac{{56}}{{8 \times {2^7}}} \cr & = \frac{7}{{128}} \cr} $$

Consider the following events :
$$A :$$  Father has at least one boy
$$B :$$  Father has $$2$$ boys and one girl
Then, $$A =$$  one boy and $$2$$ girls, $$2$$ boys and one girl, $$3$$ boys and no girl $$A \cap B = 2$$   boys and one girl
Now, the required probability is
$$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = \frac{1}{3}.$$

$$\eqalign{ & \,\,\,\,\,\,\,P\left( {\overline {A\,} \overline B } \right) \cr & = P\left( {\overline {A \cup B\,} } \right) \cr & = 1 - P\left( {A \cup B} \right) \cr & = 1 - \left\{ {P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)} \right\} \cr & = 1 - \left\{ {0.25 + 0.50 - 0.14} \right\} \cr & = 0.39 \cr} $$

The probability of hitting in one shot $$ = \frac{{10}}{{100}} = \frac{1}{{10}}$$
 If he fires $$n$$ shots, the probability of hitting at least once
$$\eqalign{ & = 1 - {\left( {1 - \frac{1}{{10}}} \right)^n} = 1 - {\left( {\frac{9}{{10}}} \right)^n} = \frac{1}{2}\,\,\left( {{\text{from the question}}} \right) \cr & \therefore \,{\left( {\frac{9}{{10}}} \right)^n} = \frac{1}{2} \cr & \therefore \,n\left\{ {2{{\log }_{10}}3 - 1} \right\} = - {\log _{10}}2 \cr & \therefore \,n = \frac{{{{\log }_{10}}2}}{{1 - 2{{\log }_{10}}3}} = \frac{{0.3010}}{{1 - 2 \times 0.4771}} = 6.5\left( {{\text{nearly}}} \right) \cr} $$
$$\therefore $$  for $$6$$ shots, the probability is about $$53\% $$  while for $$7$$ shots it is nearly $$48\% .$$

The probability of the four cards being spades $$ = \frac{{{}^{13}{C_2}}}{{{}^{52}{C_2}}} \times \frac{{{}^{11}{C_2}}}{{{}^{50}{C_2}}}.$$
Similarly, for other suits.
$$\therefore $$  the required probability $$ = 4 \times \frac{{{}^{13}{C_2} \times {}^{11}{C_2}}}{{{}^{52}{C_2} \times {}^{50}{C_2}}} = \frac{{44}}{{85 \times 49}}.$$

Given sample space = {1, 2, 3, . . . . . ,8}
Let Event
$$A$$ : Maximum of three numbers is 6.
$$B$$ : Minimum of three numbers is 3.
This is the case of conditional probability
We have to find $$P$$ (minimum) is 3 when it is given that $$P$$ (maximum) is 6.
$$\eqalign{ & \therefore \,\,P\left( {\frac{B}{A}} \right) = \frac{{P\left( {B \cap A} \right)}}{{P\left( A \right)}} \cr & = \frac{{\frac{{^2{C_1}}}{{^8{C_3}}}}}{{\frac{{^5{C_2}}}{{^8{C_3}}}}} \cr & = \frac{{^2{C_1}}}{{^5{C_2}}} \cr & = \frac{2}{{10}} \cr & = \frac{1}{5} \cr} $$

$$n\left( S \right) = 6 \times 6 \times 6$$
$$n\left( E \right) = $$   the number of solutions of $$x + y + z = 15,$$    where $$1 \leqslant x \leqslant 6,\,1 \leqslant y \leqslant 6,\,1 \leqslant z \leqslant 6$$
$$\eqalign{ & = {\text{ coefficient of }}{x^{15}}{\text{ in }}{\left( {x + {x^2} + ......{x^6}} \right)^3} \cr & = {\text{ coefficient of }}{x^{12}}{\text{ in }}{\left( {1 + x + ......{x^5}} \right)^3} \cr & = {\text{ coefficient of }}{x^{12}}{\text{ in }}{\left( {\frac{{1 - {x^6}}}{{1 - x}}} \right)^3} \cr & = {\text{ coefficient of }}{x^{12}}{\text{ in }}\left( {1 - 3.{x^6} + 3.{x^{12}} - {x^{18}}} \right).{\left( {1 - x} \right)^{ - 3}} \cr & = {\text{ coefficient of }}{x^{12}}{\text{ in }}\left( {1 - 3{x^6} + 3{x^{12}} - {x^{18}}} \right)\left( {{}^2{C_0} + {}^3{C_1}x + {}^4{C_2}{x^2} + .....} \right) \cr & = {}^{14}{C_{12}} - 3 \times {}^8{C_6} + 3 \times {}^2{C_0} = 10 \cr & \therefore \,\,P\left( E \right) = \frac{{10}}{{6 \times 6 \times 6}} = \frac{5}{{108}}. \cr} $$

Let $$P\left( A \right) = $$  probability that a randomly selected student likes tea $$= 0.3.$$
Let $$P\left( {{A_2}} \right) = $$  probability that a randomly selected student does not like tea $$ = 1 - 0.3 = 0.7.$$
Let $$P\left( B \right) = $$  probability that a randomly selected student likes coffee $$= 0.2.$$
$$\eqalign{ & \therefore \,P\left( {{A_2} \cap B} \right) \cr & = P\left( B \right) - P\left( {A \cap B} \right) \cr & = 0.2 - 0.1 \cr & = 0.1 \cr} $$
Now we have to find
$$P\left( {\frac{{{A_2}}}{B}} \right) = \frac{{P\left( {{A_2} \cap B} \right)}}{{P\left( B \right)}} = \frac{{0.1}}{{0.2}} = \frac{1}{2}$$

The total number of subsets of $$X$$ is $${2^k}.$$ So, $$n\left( S \right) = {}^{{2^k}}{C_2}.$$
$$n\left( E \right) = $$   the number of selections of two nonintersecting subsets whose union is $$X$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {{}^k{C_0} + {}^k{C_1} + {}^k{C_2} + .....} \right)$$
( $$\because $$  the number of selections in which one subset has $$r$$ elements and the rest are in the other subset $$ = {}^k{C_r}$$  and every selection appears twice in the total number of selections ).
$$\therefore \,P\left( E \right) = \frac{{\left( {\frac{1}{2}} \right){{.2}^k}}}{{{}^{{2^k}}{C_2}}} = \frac{{{2^{k - 1}}}}{{\frac{{{2^k}\left( {{2^k} - 1} \right)}}{2}}} = \frac{1}{{{2^k} - 1}}.$$