Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$A =$$  the event of $$\left( {x,\,y} \right)$$  belonging to the area $$OTQPO$$
$$B =$$  the event of $$\left( {x,\,y} \right)$$  belonging to the area $$OSQRO$$
$$\eqalign{ & P\left( A \right) = \frac{{{\text{ar}}\left( {OTQPO} \right)}}{{{\text{ar}}\left( {OPQRO} \right)}} = \frac{{\int_0^1 {\sqrt x } \,dx}}{{1 \times 1}} = \left[ {\frac{2}{3}{x^{\frac{3}{2}}}} \right] = \frac{2}{3} \cr & P\left( B \right) = \frac{{{\text{ar}}\left( {OSQRO} \right)}}{{{\text{ar}}\left( {OPQRO} \right)}} = \frac{{\int_0^1 {\sqrt y } \,dy}}{{1 \times 1}} = \frac{2}{3} \cr & P\left( {A \cap B} \right) = \frac{{{\text{ar}}\left( {OTQS} \right)}}{{{\text{ar}}\left( {OPQRO} \right)}} = \frac{{\int_0^1 {\sqrt x } \,dx - \int_0^1 {{x^2}dx} }}{{1 \times 1}} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \cr & P\left( A \right) + P\left( B \right) = \frac{2}{3} + \frac{2}{3} \ne 1. \cr & {\text{So, }}A{\text{ and }}B{\text{ are not exhaustive}}{\text{.}} \cr & P\left( A \right).P\left( B \right) = \frac{2}{3}.\frac{2}{3} = \frac{4}{9} \ne P\left( {A \cap B} \right). \cr & {\text{So, }}A{\text{ and }}B{\text{ are not independent}}{\text{.}} \cr & P\left( {A \cup B} \right) = 1,\,P\left( A \right) + P\left( B \right) = \frac{2}{3} + \frac{2}{3} \ne P\left( {A \cup B} \right). \cr & {\text{So, }}A{\text{ and }}B{\text{ are not mutually exclusive}}{\text{.}} \cr} $$

Probability that machine stops working $$ = P\left( {A \cup B \cup C} \right)$$
$$\eqalign{ & \Rightarrow P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right) \cr & \Rightarrow P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( A \right)P\left( B \right) - P\left( A \right)P\left( C \right) - P\left( B \right)P\left( C \right) + P\left( A \right)P\left( B \right)P\left( C \right) \cr & \left( {\because \,A,\,B\,\& \,C{\text{ are independent events}}} \right) \cr & \Rightarrow P\left( {A \cup B \cup C} \right) = 0.02 + 0.1 + 0.05 - \left( {0.02 \times 0.1} \right) - \left( {0.02 \times 0.05} \right) - \left( {0.1 \times 0.05} \right) + \left( {0.02 \times 0.05 \times 0.1} \right) \cr & \Rightarrow P\left( {A \cup B \cup C} \right) = 0.16 \cr} $$
$$\therefore $$  Probability that the machine will not stop working
$$\eqalign{ & = 1 - P\left( {A \cup B \cup C} \right) \cr & = 1 - 0.16 \cr & = 0.84 \cr} $$

Probability of getting a head on tossing a coin $$\left( {{P_1}} \right) = \frac{1}{2}.$$
Probability of getting a six on rolling a dice $$\left( {{P_2}} \right) = \frac{1}{6}.$$
These two events are independent.
So the probability that the coin shows the head and the dice shows $$6$$ is given by
$$P = {P_1} \times {P_2} = \frac{1}{2} \times \frac{1}{6} = \frac{1}{{12}}.$$

$$n\left( S \right) = {}^{100}{C_3}$$
As the AM of three numbers is $$25,$$  their sum $$ = 75.$$
$$\therefore \,n\left( E \right) = $$   the number of integral solutions of $${x_1} + {x_2} + {x_3} = 75,$$     where $${x_1} \geqslant 1,\,{x_2} \geqslant 1,\,{x_3} \geqslant 1$$
$$\eqalign{ & = {\text{coefficient of }}{x^{75}}{\text{ in}}{\left( {x + {x^2} + {x^3} + .....} \right)^3} \cr & {\text{ = coefficient of }}{x^{72}}{\text{ in}}{\left( {\frac{1}{{1 - x}}} \right)^3} \cr & {\text{ = coefficient of }}{x^{72}}{\text{ in}}{\left( {1 - x} \right)^{ - 3}} \cr & = {}^{74}{C_{72}} \cr & \therefore \,\,P\left( E \right) = \frac{{{}^{74}{C_{72}}}}{{{}^{100}{C_3}}} = \frac{{{}^{74}{C_{72}}}}{{{}^{100}{C_{97}}}}. \cr} $$

$$\eqalign{ & {r_1},\,{r_2},\,{r_3}\, \in \,\left\{ {1,\,2,\,3,\,4,\,5,\,6} \right\} \cr & {r_1},\,{r_2},\,{r_3}\,{\text{are of the form }}3k,\,3k + 1,\,3k + 2 \cr & {\text{Required probability}} \cr & = \frac{{3! \times {}^2{C_1} \times {}^2{C_1} \times {}^2{C_1}}}{{6 \times 6 \times 6}} \cr & = \frac{{6 \times 8}}{{216}} \cr & = \frac{2}{9} \cr} $$

If $$\omega $$ is a complex cube root of unity then, we know that
$${\omega ^{3m}} + {\omega ^{3n + 1}} + {\omega ^{3p + 2}} = 0$$
where $$m, n, p$$  are integers.
$$\therefore \,\,{r_1},{r_2},{r_3}$$   should be of the form $$3m, 3n + 1$$   and $$3p + 2$$  taken in any order. As $${r_1},{r_2},{r_3}$$   are the numbers obtained on die, these can take any value from 1 to 6.
∴ $$m$$ can take values 1 or 2, $$n$$ can take values 0 or 1
$$p$$ can take values 0 or 1
∴ Number of ways of selecting $${r_1},{r_2},{r_3}$$
$$ = {\,^2}{C_1} \times {\,^2}{C_1} \times {\,^2}{C_1} \times 3!.$$
Also the total number of ways of getting $${r_1},{r_2},{r_3}$$   on die $$ = 6 \times 6 \times 6$$
∴ Required probability $$ = \frac{{^2{C_1} \times {\,^2}{C_1} \times {\,^2}{C_1} \times 3!}}{{6 \times 6 \times 6}}$$
$$ = \frac{2}{9}$$

The probability of not drawing the ace in the first draw, in the second draw and in the third draw are $$\frac{{12}}{{13}},\,\frac{{11}}{{12}}$$  and $$\frac{{10}}{{11}}$$ respectively.
The probability of drawing ace of spades in the fourth draw $$ = \frac{1}{{10}}$$
 $$\therefore $$  the required probability $$ = \frac{{12}}{{13}} \times \frac{{11}}{{12}} \times \frac{{10}}{{11}} \times \frac{1}{{10}} = \frac{1}{{13}}$$

According to given condition, we can have the following cases
(I) GGBBB
(II) BGGBB
(lll) GBGBB
(IV) BGBGB
(V) GBBGB
i.e., the two girls can occupy two of the first three places (case I, II, III) or second and fourth (case IV) or first and fourth (case V) places.
Thus favourable cases are $$ = 3 \times 2!\, \times 3!\, + 2 \times 2!\, \times 3! = 60$$
Total ways in which 5 persons can be seated $$ = 5!\, = 120$$
∴ Required probability $$ = \frac{{60}}{{120}} = \frac{1}{2}$$

Let $$n$$ be the number of tosses and $$X$$ the number of times heads occurs. Then $$X \sim B\left( {n,\,p} \right),$$   with $$p = \frac{1}{2}.$$
Therefore, since $$P\left( {X = 7} \right) = P\left( {X = 9} \right),$$     we have
$$\eqalign{ & {}^n{C_7}{\left( {\frac{1}{2}} \right)^7}{\left( {\frac{1}{2}} \right)^{n - 7}} = {}^n{C_9}{\left( {\frac{1}{2}} \right)^9}{\left( {\frac{1}{2}} \right)^{n - 9}} \cr & \Rightarrow {}^n{C_7}{\left( {\frac{1}{2}} \right)^n} = {}^n{C_9}{\left( {\frac{1}{2}} \right)^n} \cr & {\text{That is, }}{}^n{C_7} = {}^n{C_9} = {}^n{C_{n - 9}}, \cr & {\text{yielding}}\,7 = n - 9{\text{ or }}n = 16 \cr & {\text{Hence,}}\,\,P\left( {X = 2} \right) = {}^{16}{C_2}{\left( {\frac{1}{2}} \right)^{16}} = \left( {\frac{{16 \times 15}}{2}} \right){\left( {\frac{1}{2}} \right)^{16}} = \frac{{15}}{{{2^{13}}}} \cr} $$

$$\eqalign{ & P\left( {{\text{Six occurs}}} \right) = P\left( {{S_1}} \right) = \frac{1}{6} \cr & P\left( {{\text{Six does not occur}}} \right) = P\left( {{S_2}} \right) = \frac{5}{6} \cr & P\left( {E|{S_1}} \right) = P\left( {{\text{man speaks truth}}} \right) = \frac{3}{4}{\text{ }} \cr & P\left( {E|{S_2}} \right) = P\left( {{\text{man does not speak the truth}}} \right) = \frac{1}{4} \cr & \therefore \,{\text{By Baye's theorem,}} \cr & P\left( {{S_1}|E} \right) = P\left( {{\text{he reports that six has occured is actually a six}}} \right) \cr & {\text{ = }}\frac{{P\left( {{S_1}} \right)P\left( {E|{S_1}} \right)}}{{P\left( {{S_1}} \right)P\left( {E|{S_1}} \right) + P\left( {{S_2}} \right)P\left( {E|{S_2}} \right)}} \cr & {\text{ = }}\frac{{\frac{1}{6} \times \frac{3}{4}}}{{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}}} \cr & {\text{ = }}\frac{3}{8}{\text{ }} \cr} $$