Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$\eqalign{ & P = {}^5{C_2}{\left( {\frac{1}{2}} \right)^2}{\left( {\frac{1}{2}} \right)^3} + {}^5{C_3}{\left( {\frac{1}{2}} \right)^3}{\left( {\frac{1}{2}} \right)^2} + {}^5{C_4}{\left( {\frac{1}{2}} \right)^4}{\left( {\frac{1}{2}} \right)^1} + {}^5{C_5}{\left( {\frac{1}{2}} \right)^5}{\left( {\frac{1}{2}} \right)^0} \cr & = {\left( {\frac{1}{2}} \right)^5}\left[ {{}^5{C_2} + {}^5{C_3} + {}^5{C_4} + {}^5{C_5}} \right] \cr & = \frac{1}{{{3^2}}}\left[ {10 + 10 + 5 + 1} \right] \cr & = \frac{1}{{{3^2}}} \times 26 \cr & = \frac{{13}}{{16}} \cr} $$

$$\eqalign{ & P\left( {{E_1}} \right) = \frac{1}{2},\,P\left( {{E_2}} \right) = \frac{1}{3}{\text{ and }}P\left( {{E_3}} \right) = \frac{1}{4}; \cr & P\left( {{E_1}U{E_2}U{E_3}} \right) = 1 - P\left( {{{\overline E }_1}} \right)P\left( {{{\overline E }_2}} \right)P\left( {{{\overline E }_3}} \right) \cr & = 1 - \left( {1 - \frac{1}{2}} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right) \cr & = \frac{3}{4} \cr} $$

Here number of errors per page $$ = \frac{{250}}{{500}} = \frac{1}{2}$$   and $$n = 2$$
$$\therefore \,\lambda = np = 2 \times \frac{1}{2} = 1$$
and probability of no error
$$P\left( {X = 0} \right) = \frac{{{e^{ - 1}} \times {{\left( 1 \right)}^0}}}{{0!}} = {e^{ - 1}}$$

Let $${A_i}\left( {i = 2,\,3,\,4,\,5} \right)$$     be the event that urn contains $$2,\,3,\,4,\,5$$    white balls and let $$B$$ be the event that two white balls have been drawn then we have to find $$P\left( {\frac{{{A_5}}}{B}} \right).$$
Since the four events $${A_2},\,{A_3},\,{A_4}$$   and $${A_5}$$ are equally likely we have $$P\left( {{A_2}} \right) = P\left( {{A_3}} \right) = P\left( {{A_4}} \right) = P\left( {{A_5}} \right) = \frac{1}{4}.$$
$$P\left( {\frac{B}{{{A_2}}}} \right)$$  is probability of event that the urn contains $$2$$ white balls and both have been drawn.
$$\eqalign{ & \therefore \,P\left( {\frac{B}{{{A_2}}}} \right) = \frac{{{}^2{C_2}}}{{{}^5{C_2}}} = \frac{1}{{10}} \cr & {\text{Similarly,}} \cr & P\left( {\frac{B}{{{A_3}}}} \right) = \frac{{{}^3{C_2}}}{{{}^5{C_2}}} = \frac{3}{{10}}, \cr & P\left( {\frac{B}{{{A_4}}}} \right) = \frac{{{}^4{C_2}}}{{{}^5{C_2}}} = \frac{3}{5}, \cr & P\left( {\frac{B}{{{A_5}}}} \right) = \frac{{{}^5{C_2}}}{{{}^5{C_2}}} = 1 \cr & {\text{By Bayes theorem,}} \cr & P\left( {\frac{{{A_5}}}{B}} \right) = \frac{{P\left( {{A_5}} \right)P\left( {\frac{B}{{{A_5}}}} \right)}}{{\left( {P\left( {{A_2}} \right)P\left( {\frac{B}{{{A_2}}}} \right) + P\left( {{A_3}} \right)P\left( {\frac{B}{{{A_3}}}} \right) + P\left( {{A_4}} \right)P\left( {\frac{B}{{{A_4}}}} \right) + P\left( {{A_5}} \right)P\left( {\frac{B}{{{A_5}}}} \right)} \right)}} \cr & = \frac{{\frac{1}{4}.1}}{{\frac{1}{4}\left[ {\frac{1}{{10}} + \frac{3}{{10}} + \frac{3}{5} + 1} \right]}} \cr & = \frac{{10}}{{20}} \cr & = \frac{1}{2} \cr} $$

The probability of both drawing the common card $$x$$
$$ = P\left( x \right) = $$   (probability of $$A$$ drawing the card $$x$$ and any other card $$y$$) × (probability of $$B$$ drawing the card $$x$$ and a card other than $$y$$)
$$\therefore \,P\left( x \right) = \frac{{{}^{51}{C_1}}}{{{}^{52}{C_2}}} \times \frac{{{}^{50}{C_1}}}{{{}^{52}{C_2}}}$$     for all $$x,$$ where $$x$$ has $$52$$  values.
$$\therefore $$  the required probability $$ = \sum {P\left( x \right)} = 52 \times \frac{{51 \times 50 \times 4}}{{52 \times 51 \times 52 \times 51}} = \frac{{50}}{{663}}.$$

When a die is thrown, the sample space $$\left( S \right)$$ is
$$S = \left\{ {1,\,2,\,3,\,4,\,5,\,6} \right\}$$
Let $$A :$$  the number is even $$ = \left\{ {2,\,4,\,6} \right\}$$
$$\therefore \,P\left( A \right) = \frac{3}{6} = \frac{1}{2}$$
$$B :$$  the number is red $$ = \left\{ {1,\,2,\,3} \right\}$$
$$\eqalign{ & \therefore \,P\left( B \right) = \frac{3}{6} = \frac{1}{2} \cr & A \cap B = \left\{ 2 \right\},\,P\left( {A \cap B} \right) = \frac{1}{6} \cr & {\text{or }}P\left( A \right).P\left( B \right) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \ne \frac{1}{6} \cr} $$

Given that $$P\left( B \right) = \frac{3}{4},P\left( {A \cap B \cap \overline C } \right) = \frac{1}{3}$$
$$P\left( {\overline A \cap B \cap \overline C } \right) = \frac{1}{3}$$
From venn diagram, we see

$$\eqalign{ & B \cap C \equiv B - \left( {A \cap B \cap \overline C } \right) - \left( {\overline A \cap B \cap \overline C } \right) \cr & \Rightarrow P\left( {B \cap C} \right) = P\left( B \right) - P\left( {A \cap B \cap \overline C } \right) - P\left( {\overline A \cap B \cap \overline C } \right) \cr & \Rightarrow \,\,P\left( {B \cap C} \right) = \frac{3}{4} - \frac{1}{3} - \frac{1}{3} \cr & = \frac{{9 - 4 - 4}}{{12}} \cr & = \frac{1}{{12}} \cr} $$

The probability of getting at least $$3$$ in a throw $$ = \frac{4}{6} = \frac{2}{3}$$
$$\therefore $$  the required probability
$$ = {}^6{C_3}.{\left( {\frac{2}{3}} \right)^3}.{\left( {\frac{1}{3}} \right)^3} + {}^6{C_4}.{\left( {\frac{2}{3}} \right)^4}.{\left( {\frac{1}{3}} \right)^2} + {}^6{C_5}.{\left( {\frac{2}{3}} \right)^5}.\frac{1}{3} + {}^6{C_6}.{\left( {\frac{2}{3}} \right)^6}$$

$$\eqalign{ & P\left( {\frac{{\overline A }}{{\overline B }}} \right) = \frac{{P\left( {\overline A \cap \overline B } \right)}}{{P\left( {\overline B } \right)}} \cr & = \frac{{P\left( {\overline {A \cup B} } \right)}}{{P\left( {\overline B } \right)}} \cr & = \frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}} \cr} $$

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