Probability MCQ Questions & Answers in Statistics and Probability | Maths

Let $$A$$ $$ \equiv $$ Sum of the digits is 8
$$B$$ $$ \equiv $$ Product of the digits is 0
(Then $$A$$ = {08, 17, 26, 35, 44}
$$B$$ = {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40,}
$$\eqalign{ & A \cap B = \left\{ {08} \right\} \cr & \therefore \,\,P\left( {\frac{A}{B}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} \cr & = \frac{{\frac{1}{50}}}{{\frac{{14}}{{50}}}} \cr & = \frac{1}{{14}} \cr} $$

$$\eqalign{ & {E_1}\,:{\text{coin is fair, }}{E_2}\,:{\text{coin is biased,}} \cr & A{\text{ second toss shows tail}}{\text{.}} \cr & P\left( {\frac{{{E_1}}}{A}} \right) = \frac{{P\left( {\frac{A}{{{E_1}}}} \right)P\left( {{E_1}} \right)}}{{P\left( {\frac{A}{{{E_1}}}} \right)P\left( {{E_1}} \right) + P\left( {\frac{A}{{{E_2}}}} \right)P\left( {{E_2}} \right)}} \cr & = \frac{{\frac{m}{N}.\frac{1}{2}.\frac{1}{2}}}{{\frac{m}{N}.\frac{1}{2}.\frac{1}{2} + \frac{{N - m}}{N}.\frac{2}{3}.\frac{1}{3}}} \cr & = \frac{{9m}}{{8N + m}} \cr} $$

As $$0.4 + 0.6 = 1,$$    the man either takes a step forward or a step backward. Let a step forward be a success and a step backward be a failure.
Then, the probability of success in one step $$ = p = 0.4 = \frac{2}{5}$$
The probability of failure in one step $$ = q = 0.6 = \frac{3}{5}$$
In $$11$$  steps he will be one step away from the starting point if the numbers of successes and failures differ by $$1.$$
So, the number of successes $$ = 6.$$  The number of failures $$ = 5.$$
or the number of successes $$ = 5.$$  The number of failures $$= 6.$$
$$\therefore $$  The required probability
$$\eqalign{ & = {}^{11}{C_6}{p^6}{q^5} + {}^{11}{C_5}{p^5}{q^6} \cr & = {}^{11}{C_6}{\left( {\frac{2}{5}} \right)^6}.{\left( {\frac{3}{5}} \right)^5} + {}^{11}{C_5}{\left( {\frac{2}{5}} \right)^5}.{\left( {\frac{3}{5}} \right)^6} \cr & = 462 \times {\left( {\frac{6}{{25}}} \right)^5} \cr} $$

Probability of all the letters kept in the right envelope is
$$\frac{1}{{n!}}\left( {\because \,{\text{Total letters}} = n} \right){\text{i}}{\text{.e}}{\text{., }}P = \frac{1}{{n!}}$$
We know, if $$q$$ is the term used for the probability of the letters which are not kept in the right envelope.
Then, $$p + q = 1 \Rightarrow q = 1 - p = 1 - \frac{1}{{n!}}$$

$$P\left( {{E_n}} \right) = $$   probability of the dice showing $$n = kn.$$
Clearly, $$P\left( {{E_1}} \right) + P\left( {{E_2}} \right) + ..... + P\left( {{E_6}} \right) = 1$$
$$\eqalign{ & \therefore \,k\left( {1 + 2 + 3 + 4 + 5} \right) = 1 \cr & \therefore \,k = \frac{1}{{\frac{{6.7}}{2}}} = \frac{1}{{21}} \cr & {\text{So, }}P\left( {{E_3}} \right) = k.3 = \frac{1}{{21}} \times 3 = \frac{1}{7}. \cr} $$

$$\eqalign{ & {\text{We know that }}P{\text{ }}\left( {{\text{exactly one of }}A{\text{ or }}B{\text{ occurs}}} \right) \cr & = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) \cr & \therefore \,P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) = p......\left( 1 \right) \cr & {\text{Similarly,}} \cr & P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) = p......\left( 2 \right) \cr & {\text{and }}P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) = p......\left( 3 \right) \cr & {\text{Adding equation}}\left( 1 \right),\,\left( 2 \right){\text{ and }}\left( 3 \right){\text{ we get}} \cr & 2\left[ {P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right)} \right] = 3p \cr & \Rightarrow \left[ {P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right)} \right] = \frac{{3p}}{2}.....\left( 4 \right) \cr & {\text{It is also given that}} \cr & P\left( {A \cap B \cap C} \right) = {p^2}......\left( 5 \right) \cr & {\text{Now,}} \cr & P\left( {{\text{at least one of }}A,{\text{ }}B{\text{ and }}C} \right) \cr & = p\left( A \right) + p\left( B \right) + p\left( C \right) - p\left( {A \cap B} \right) - p\left( {B \cap C} \right) - p\left( {C \cap A} \right) + p\left( {A \cap B \cap C} \right) \cr & = \frac{{3p}}{2} + {p^2} \cr & = \frac{{3p + 2{p^2}}}{2}\,\,\,\,\,\left[ {{\text{Using equation }}\left( 4 \right)\,{\text{and}}\left( 5 \right)} \right] \cr} $$

A pair of fair dice is thrown, the sample space $$S$$ = (1, 1) (1,2) (1, 3) . . . . = 36
Possibility of getting 9 are (5,4) , (4, 5), (6,3), (3,6)
∴ Probability of getting score 9 in a single throw
$$\eqalign{ & = \frac{4}{{36}} \cr & = \frac{1}{9} \cr} $$
∴ Probability of getting score 9 exactly twice
$$\eqalign{ & = {\,^3}{C_2} \times {\left( {\frac{1}{9}} \right)^2}.\left( {1 - \frac{1}{9}} \right) \cr & = \frac{{3!}}{{2!}} \times \frac{1}{9} \times \frac{1}{9} \times \frac{8}{9} \cr & = \frac{{3.2!}}{{2!}} \times \frac{1}{9} \times \frac{1}{9} \times \frac{8}{9} \cr & = \frac{8}{{243}} \cr} $$

Total number of selecting $$3$$ components out of $$10 = {}^{10}{C_3}.$$
Out of $$3$$ selected components two defective pieces can be selected in $${}^4{C_2}$$ ways and one non-defective piece will be selected in $${}^6{C_1}$$ ways.
Hence, required probability is
$$ = \frac{{{}^6{C_1} \times {}^4{C_2}}}{{{}^{10}{C_3}}} = \frac{{6 \times 6 \times 6}}{{10 \times 9 \times 8}} = \frac{3}{{10}}$$

Probability that $$\left( {X = 3} \right) = \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{4} = \frac{7}{{24}}$$

$$\eqalign{ & n\left( S \right) = {}^{50}{C_5}, \cr & n\left( E \right) = {}^{30}{C_2} \times {}^{19}{C_2} \cr & \therefore \,P\left( E \right) = \frac{{{}^{30}{C_2} \times {}^{19}{C_2}}}{{{}^{50}{C_5}}} \cr} $$