Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$\eqalign{ & {\text{We have,}} \cr & P\left( {\left| {X - 4} \right| \leqslant 2} \right) \cr & = P\left( { - 2 \leqslant X - 4 \leqslant 2} \right) \cr & = P\left( {2 \leqslant X \leqslant 6} \right) \cr & = 1 - \left[ {P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 7} \right) + P\left( {X = 8} \right)} \right] \cr & = 1 - \left[ {{}^8{C_0}{{\left( {\frac{1}{2}} \right)}^8} + {}^8{C_1}{{\left( {\frac{1}{2}} \right)}^8} + {}^8{C_7}{{\left( {\frac{1}{2}} \right)}^8} + {}^8{C_8}{{\left( {\frac{1}{2}} \right)}^8}} \right] \cr & = 1 - {\left( {\frac{1}{2}} \right)^8}\left( {1 + 8 + 8 + 1} \right) \cr & = 1 - \frac{{18}}{{{2^8}}} \cr & = \frac{{119}}{{128}} \cr} $$

$$\eqalign{ & P\left( {{E_1}} \right) = \frac{1}{6} \cr & P\left( {{E_2}} \right) = \frac{1}{6} \cr & P\left( {{E_3}} \right) = \frac{{2 + 4 + 6 + 4 + 2}}{{36}} = \frac{1}{2} \cr & P\left( {{E_2} \cap {E_3}} \right) = \frac{1}{2} \times \frac{1}{6} = P\left( {{E_2}} \right) \times P\left( {{E_3}} \right) \cr & P\left( {{E_1} \cap {E_3}} \right) = \frac{1}{2} \times \frac{1}{6} = P\left( {{E_1}} \right) \times P\left( {{E_3}} \right) \cr & P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = 0 \cr & P\left( {{E_1}} \right)P\left( {{E_2}} \right)P\left( {{E_3}} \right) = 0 \cr & \therefore \,P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = P\left( {{E_1}} \right)P\left( {{E_2}} \right)P\left( {{E_3}} \right) \cr & {E_1},\,{E_3}{\text{ are independent}} \cr & {E_2},\,{E_3}{\text{ are independent}} \cr} $$

Probability of $$A$$ winning [$$A$$ can win in $${1^{st}}$$ or $${3^{rd}}$$ or $${5^{th}}$$.... games if $$B$$ loses $${2^{nd}}$$ or $${4^{th}}$$ or... games]
$$\eqalign{ & = \frac{p}{{p + q}} + {\left( {\frac{q}{{p + q}}} \right)^2}.\frac{p}{{p + q}} + {\left( {\frac{q}{{p + q}}} \right)^4}.\frac{p}{{p + q}} + ..... \cr & = \frac{{\frac{p}{{p + q}}}}{{1 - {{\left( {\frac{q}{{p + q}}} \right)}^2}}}\,\,\,\,\,\left[ {{\text{In infinite G}}{\text{.P}}{\text{.S}}{\text{.}} = \frac{a}{{1 - r}}} \right] \cr & = \frac{{p\left( {p + q} \right)}}{{{{\left( {p + q} \right)}^2} - {q^2}}} \cr & {\text{Probability of }}B{\text{ winning}} \cr & = 1 - \frac{{p\left( {p + q} \right)}}{{{{\left( {p + q} \right)}^2} - {q^2}}} \cr & = \frac{{{{\left( {p + q} \right)}^2} - {q^2} - p\left( {p + q} \right)}}{{{{\left( {p + q} \right)}^2} - {q^2}}} \cr & {\text{Given, }}P\left( A \right) = 3P\left( B \right) \cr & \Rightarrow p\left( {p + q} \right) = 3\left[ {{{\left( {p + q} \right)}^2} - {q^2} - p\left( {p + q} \right)} \right] \cr & \Rightarrow 4p\left( {p + q} \right) = 3\left( {p + 2q} \right).p \cr & \Rightarrow 4p + 4q = 3p + 6q \cr & \Rightarrow p = 2q \cr & \Rightarrow \frac{p}{q} = 2{\text{ or }}p:q = 2:1 \cr} $$

The man has to win at least $$4$$ times.
$$\therefore $$  the required probability
$$\eqalign{ & = {}^7{C_4}.{\left( {\frac{1}{2}} \right)^4}.{\left( {\frac{1}{2}} \right)^3} + {}^7{C_5}.{\left( {\frac{1}{2}} \right)^5}.{\left( {\frac{1}{2}} \right)^2} + {}^7{C_6}.{\left( {\frac{1}{2}} \right)^6}.\frac{1}{2} + {}^7{C_7}{\left( {\frac{1}{2}} \right)^7} \cr & = \left( {{}^7{C_4} + {}^7{C_5} + {}^7{C_6} + {}^7{C_7}} \right).\frac{1}{{{2^7}}} \cr & = \frac{{64}}{{{2^7}}} \cr & = \frac{1}{2} \cr} $$

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

There are two cases when Abhay will say 'No'
$${\bf{Case }}\left( {\bf{i}} \right)\,{\bf{:}}$$   The number that came out is a prime and Abhay is not speaking truth, probability for this case is $$P\left( P \right) \times P\left( {{T^{^,}}} \right).$$
Here $$P\left( P \right) = $$  probability of getting a prime $$ = \frac{3}{6} = \frac{1}{2} = 0.5.$$
$$P\left( T \right)$$  is probability that Abhay is speaking truth and $$P\left( T \right)$$
$$ = 0.6{\text{ so }}P\left( {{T^{^,}}} \right) = 0.4.$$
So probability for this case is $$0.5 \times 0.4 = 0.2.$$
$${\bf{Case }}\left( {{\bf{ii}}} \right)\,{\bf{:}}$$   The number that came out is not a prime and Abhay is speaking truth, probability for this case is
$$P\left( {{P^,}} \right) \times P\left( T \right) = 0.5 \times 0.6 = 0.3.$$
So total probability for the given case is $$0.3 + 0.2 = 0.5.$$
New sample space is $$0.5$$  and we have to find the probability of case $$\left( {\bf{i}} \right)$$ which is $$\frac{{0.2}}{{0.5}} = 0.4.$$

Total number of ways in which $$4$$ persons can be selected out of $$3 + 2 + 4 = 9$$    persons $$ = {}^9{C_4} = 126$$
Number of ways in which a selection of $$4$$ contains exactly $$2$$ children $$ = {}^4{C_2} \times {}^5{C_2} = 60$$
$$\therefore $$  Required probability $$ = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$

Probability of no one hitting the target
$$ = \frac{1}{{5 \times 4 \times 3}} = \frac{1}{{60}}$$
Probability of one hitting the target
$$ = \frac{{4 + 3 + 2}}{{60}} = \frac{9}{{60}}$$
$$\therefore $$  Probability of maximum one hit
$$ = \frac{1}{{60}} + \frac{9}{{60}} = \frac{{10}}{{60}} = \frac{1}{6}$$
Probability that two shots are hit at least is the required probability
$$ = 1 - \frac{1}{6} = \frac{5}{6}$$

Required probability $$ = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$

Given limit
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} \right)^{\frac{2}{x}}} \cr & = \mathop {\lim }\limits_{x \to 0} {\left( {1 + \frac{{{a^x} + {b^x} - 2}}{2}} \right)^{\frac{2}{{{a^x} + {b^x} - 2}}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} - 1 + {b^x} - 1}}{x}} \right)}} \cr & = {e^{\log \,ab}} \cr & = ab \cr & = 6 \cr} $$
Total number of possible ways in which $$a,\,b$$  can take values is $$6 \times 6 = 36.$$
Total possible ways are $$\left( {1,\,6} \right),\,\left( {6,\,1} \right),\,\left( {2,\,3} \right),\,\left( {3,\,2} \right).$$
The total number of possible ways is $$4.$$
Hence, the required probability is $$\frac{4}{{36}} = \frac{1}{9}.$$