Probability MCQ Questions & Answers in Statistics and Probability | Maths

Let the two events be $${E_1}$$ and $${E_2}$$
Let their chances be $$p$$ and $$q$$ respectively.
Then $$p = {q^2}......\left( {\text{i}} \right)$$
The chances of not happening of the events are $$1 - p$$  and $$1 - q$$  respectively.
Odds against the first event $$ = \frac{{1 - p}}{p}$$
Odds against the second event $$ = \frac{{1 - q}}{q}$$
$$\eqalign{ & {\text{Given,}} \cr & \frac{{1 - p}}{p} = {\left( {\frac{{1 - q}}{q}} \right)^3} \cr & \Rightarrow \frac{{1 - {q^2}}}{{{q^2}}} = \frac{{{{\left( {1 - q} \right)}^3}}}{{{q^3}}} \cr & \left[ {{\text{From eqution }}\left( {\text{i}} \right)} \right] \cr & \Rightarrow \left( {\frac{{1 - q}}{{{q^2}}}} \right)\left[ {\left( {1 + q} \right) - \frac{{{{\left( {1 - q} \right)}^2}}}{q}} \right] = 0 \cr & \because \,q \ne 1{\text{ and }}q \ne 0 \cr & \therefore \,q\left( {1 + q} \right) = 1 - 2q + {q^2} \Rightarrow q = \frac{1}{3} \cr & \therefore {\text{ from equation }}\left( {\text{i}} \right)\,p = {q^2} = \frac{1}{9} \cr & \therefore \,p\left( {{E_1}} \right) = p = \frac{1}{9} \cr} $$

$$\eqalign{ & {\text{Favourable ways}} = 3!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}}} \right) = 2 \cr & {\text{Total ways}} = 3! \cr & \therefore \,{\text{Required probability}} = \frac{2}{{3!}} = \frac{1}{3} \cr} $$

Let $${E_1} = $$  the event of drawing a white ball from the first box.
Similarly, $${E_2}$$ and $${E_3}$$ Here $$P\left( {{E_1}} \right) = \frac{3}{4},\,\,P\left( {{E_2}} \right) = \frac{1}{2},\,\,P\left( {{E_3}} \right) = \frac{1}{4}$$
The required probability
$$\eqalign{ & = P\left( {{E_1}{E_2}\overline {{E_3}} } \right) + P\left( {{E_1}\overline {{E_2}} {E_3}} \right) + P\left( {\overline {{E_1}} {E_2}{E_3}} \right) \cr & = P\left( {{E_1}} \right).P\left( {{E_2}} \right).P\left( {\overline {{E_3}} } \right) + P\left( {{E_1}} \right).P\left( {\overline {{E_2}} } \right).P\left( {{E_3}} \right) + P\left( {\overline {{E_1}} } \right).P\left( {{E_2}} \right).P\left( {{E_3}} \right) \cr & = \frac{3}{4}.\frac{1}{2}.\frac{3}{4} + \frac{3}{4}.\frac{1}{2}.\frac{1}{4} + \frac{1}{4}.\frac{1}{2}.\frac{1}{4} = \frac{{13}}{{32}}. \cr} $$

The event follows binomial distribution with
$$\eqalign{ & n = 5,p = \frac{3}{6} = \frac{1}{2}. \cr & q = 1 - p = \frac{1}{2}.; \cr & \therefore \,\,{\text{Variance}} = npq = \frac{5}{4}. \cr} $$

Total time of a cycle $$= 60\,sec$$
Time of green light $$= 25\,sec$$
$$\therefore $$  Probability that light is green $$ = \frac{{25}}{{60}}$$
$$\therefore $$  Probability that light will not be green $$ = 1 - \frac{{25}}{{60}} = \frac{7}{{12}}$$

Seven people can seat themselves at a round table in $$6!$$  ways. The number of ways in which two distinguished persons will be next to each other $$ = 2\left( 5 \right)!.$$
Hence, the required probability $$ = \frac{{2\left( 5 \right)!}}{{6!}} = \frac{1}{3}.$$

$$\eqalign{ & P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr & = 0.25 + 0.50 - 0.14 = 0.61 \cr & \therefore \,\,P\left( {A' \cap B'} \right) = P\left( {\left( {A \cup B} \right)'} \right) = 1 - P\left( {A \cup B} \right) \cr & = 1 - 0.61 \cr & = 0.39 \cr} $$

A pair of fair dice is thrown, the sample space
$$S = \left( {1,\,1} \right),\,\left( {1,\,2} \right),\,\left( {1,\,3} \right)...... = 36$$
Possibility of getting $$9$$ are $$\left( {5,\,4} \right),\,\left( {4,\,5} \right),\,\left( {6,\,3} \right),\,\left( {3,\,6} \right)$$
$$\therefore $$  Possibility of getting score $$9$$ in a single throw
$$ = \frac{4}{{36}} = \frac{1}{9}$$
$$\therefore $$  Probability of getting score $$9$$ exactly twice
$$\eqalign{ & = {}^3{C_2} \times {\left( {\frac{1}{9}} \right)^2}.\left( {1 - \frac{1}{9}} \right) \cr & = \frac{{3!}}{{2!}} \times \frac{1}{9} \times \frac{1}{9} \times \frac{8}{9} \cr & = \frac{8}{{243}} \cr} $$

$$n\left( S \right) = 7!,\,n\left( E \right) = \left( {3!} \right) \times \left( {4!} \right)$$       ($$\because $$  after making $$4$$ gentlemen sit in $$3!$$  ways, $$4$$ ladies can sit in $$4!$$  ways in between the gentlemen)
$$\therefore \,P\left( E \right) = \frac{{\left( {3!} \right) \times \left( {4!} \right)}}{{7!}} = \frac{6}{{7 \times 6 \times 5}} = \frac{1}{{35}}.$$

$$E = \left\{ {\left( {1,\,6} \right),\,\left( {2,\,5} \right),\,\left( {3,\,4} \right),\,\left( {4,\,3} \right),\,\,\left( {5,\,2} \right)\left( {6,\,1} \right)} \right\}$$
$$\therefore \,P\left( E \right) = \frac{6}{{6 \times 6}} = \frac{1}{6}.$$     So, the odds against drawing $$7 = \frac{{P\left( {\overline E } \right)}}{{P\left( E \right)}} = \frac{{1 - \frac{1}{6}}}{{\frac{1}{6}}} = \frac{5}{1}.$$