Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$n\left( S \right) = $$   the area of the circle of radius $$r$$
$$n\left( E \right) = $$   the area of the circle of radius $$\frac{r}{2}$$
$$\therefore $$  The probability $$ = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{\pi {{\left( {\frac{r}{2}} \right)}^2}}}{{\pi {r^2}}} = \frac{1}{4}$$

Three vertices can be selected in $${}^6{C_3}$$ ways

The only equilateral triangles possible are $${A_1}{A_3}{A_5}$$  and $${A_2}{A_4}{A_6}$$
$$p = \frac{2}{{{}^6{C_3}}} = \frac{2}{{20}} = \frac{1}{{10}}$$

Since, India's second win occurs at the third test.
Therefore, the sample space is $$ = \left[ {LWW,\,WLW} \right]$$
where, $$L =$$  losing the test
$$W =$$  winning the test.
$$\therefore \,P$$  (India's win occur at the $${3^{rd}}$$ test)
$$\eqalign{ & = P\left( {LWW} \right) + P\left( {WLW} \right) \cr & = P\left( L \right)P\left( W \right)P\left( W \right) + P\left( W \right)P\left( L \right)P\left( W \right) \cr} $$
($$\because $$  Probability from match to match is independent).
$$\eqalign{ & = \left( {\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}} \right) + \left( {\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}} \right)\,\,\,\left( {{\text{given}}} \right) \cr & = \frac{1}{8} + \frac{1}{8} \cr & = \frac{2}{8} \cr & = \frac{1}{4} \cr} $$

Total number of arrangements of the letters of the word UNIVERSITY is $$\frac{{10!}}{{2!}}.$$
Number of arrangements when both I's are together $$ = 9!$$
So. the number of ways in which $$2$$ I’s do not together $$\frac{{10!}}{{2!}} - 9!$$
$$\therefore $$  Required probability
$$\eqalign{ & = \frac{{\frac{{10!}}{{2!}} - 9!}}{{\frac{{10!}}{{2!}}}} \cr & = \frac{{10! - 9!\,2!}}{{10!}} \cr & = \frac{{10! \times 9! - 9!.2!}}{{10!}} \cr & = \frac{{9!\left[ {10 - 2} \right]}}{{10 \times 9!}} \cr & = \frac{8}{{10}} \cr & = \frac{4}{5} \cr} $$

Total number of matches $$= n = 5$$
India will win the series if it wins either $$3$$ or $$4$$ or $$5$$ matches.
In previous question we have calculated the value of $$P\left( 3 \right) = $$   probability of winning $$3$$ matches
$$ = \left( {{}^5{C_3}} \right){\left( {\frac{2}{3}} \right)^3}{\left( {\frac{1}{3}} \right)^2}$$
Required Probability
$$\eqalign{ & = P\left( 3 \right) + P\left( 4 \right) + P\left( 5 \right) \cr & = \left( {{}^5{C_3}} \right){\left( {\frac{2}{3}} \right)^3}{\left( {\frac{1}{3}} \right)^2} + \left( {{}^5{C_4}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^1} + \left( {{}^5{C_5}} \right){\left( {\frac{2}{3}} \right)^5}{\left( {\frac{1}{3}} \right)^0} \cr & = \frac{{10'\,\,8}}{{243}} + \frac{{5'\,\,16}}{{243}} + \frac{{1'\,\,32}}{{243}} \cr & = \frac{{192}}{{243}} \cr} $$

$$\eqalign{ & P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}} \cr & \Rightarrow \frac{1}{2} = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{\frac{1}{4}}} \cr & \Rightarrow P\left( {{E_1} \cap {E_2}} \right) = \frac{1}{8} \cr & = P\left( {{E_2}} \right).P\left( {\frac{{{E_1}}}{{{E_2}}}} \right) = P\left( {{E_2}} \right).\frac{1}{4} \cr & \Rightarrow P\left( {{E_2}} \right) = \frac{1}{2} \cr & {\text{Since, }}P\left( {{E_1} \cap {E_2}} \right) = \frac{1}{8} = P\left( {{E_1}} \right).P\left( {{E_2}} \right) \cr & \Rightarrow {\text{events are independent}} \cr & {\text{Also, }}P\left( {{E_1} \cup {E_2}} \right) = \frac{1}{2} + \frac{1}{4} - \frac{1}{8} = \frac{5}{8} \cr & \Rightarrow {E_1}\,\,\& \,\,{E_2}\,{\text{are non exhaustive}}{\text{.}} \cr} $$

$$\eqalign{ & P\left( E \right) = P\left( {2\,\,{\text{or 3 or 5 or 7}}} \right) \cr & = 0.23 + 0.12 + 0.20 + 0.07 \cr & = 0.62 \cr & P\left( F \right) = P\left( {1\,{\text{or 2 or 3}}} \right) \cr & = 0.15 + 0.23 + 0.12 \cr & = 0.50 \cr & \therefore P\left( {E \cup F} \right) = P\left( E \right) + P\left( F \right) - P\left( {E \cap F} \right) \cr & = 0.62 + 0.50 - 0.35 \cr & = 0.77 \cr} $$

$$\eqalign{ & {\text{Mean}} = np = 4{\text{ and}} \cr & {\text{Variance}} = npq = 2 \cr & \therefore \,p = q = \frac{1}{2}{\text{ and }}n = 8 \cr & \therefore \,p\left( {2{\text{ success}}} \right) = {}^8{C_2}{\left( {\frac{1}{2}} \right)^6}{\left( {\frac{1}{2}} \right)^2} \cr & = \frac{{28}}{{{2^8}}} \cr & = \frac{{28}}{{256}} \cr} $$

$${E_1} = $$  the event of drawing from the first box.
$${E_2} = $$  the event of drawing from the second box.
$$P\left( {{E_1}} \right) = \frac{3}{6} = \frac{1}{2}$$    and $$P\left( {{E_2}} \right) = \frac{3}{6} = \frac{1}{2}$$
Clearly, $${E_1}$$ and $${E_2}$$ are mutually exclusive and exhaustive.
Now, $$P\left( {\frac{B}{{{E_1}}}} \right) = \frac{3}{7}$$   and $$P\left( {\frac{B}{{{E_2}}}} \right) = \frac{4}{7}$$
We have to find $$P\left( {\frac{{{E_1}}}{B}} \right)$$
By Bayes’ theorem,
$$\eqalign{ & P\left( {\frac{{{E_1}}}{B}} \right) = \frac{{P\left( {{E_1}} \right).P\left( {\frac{B}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right).P\left( {\frac{B}{{{E_1}}}} \right) + P\left( {{E_2}} \right).P\left( {\frac{B}{{{E_2}}}} \right)}} \cr & = \frac{{\frac{1}{2}.\frac{3}{7}}}{{\frac{1}{2}.\frac{3}{7} + \frac{1}{2}.\frac{4}{7}}} \cr & = \frac{3}{7} \cr} $$

He will fail in exam in two cases :
$${\bf{Case}}\,\left( {\bf{i}} \right)\,{\bf{:}}$$   He studied and failed, probability of this case is $$\left( {\frac{1}{3}} \right)\left( {\frac{1}{2}} \right) = \frac{1}{6}$$
$${\bf{Case}}\,\left( {{\bf{ii}}} \right)\,{\bf{:}}$$   He didn’t studied and failed, probability of this case is $$\left( {\frac{2}{3}} \right)\left( {\frac{3}{4}} \right) = \frac{1}{2}$$
So total probability is $$\frac{1}{6} + \frac{1}{2} = \frac{4}{6} = \frac{2}{3}$$
Then required probability $$ = \frac{{\left( {\frac{1}{2}} \right)}}{{\left( {\frac{2}{3}} \right)}} = \frac{3}{4}$$