Probability MCQ Questions & Answers in Statistics and Probability | Maths

The required probability
$$ = 1 - $$   probability of equal number of heads and tails
$$\eqalign{ & = 1 - {}^{2n}{C_n}.{\left( {\frac{1}{2}} \right)^n}.{\left( {\frac{1}{2}} \right)^{2n - n}} \cr & = 1 - \frac{{\left( {2n} \right)!}}{{\left( {n!} \right)\left( {n!} \right)}}.{\left( {\frac{1}{4}} \right)^n} \cr} $$

Lets define the events as
Probability of getting project copy $$\left( A \right) = p$$
Probability of getting blue pen $$\left( B \right) = q$$
Probability of getting black pen $$\left( C \right) = \frac{1}{2}$$
Then,
$$\eqalign{ & {\text{ }}p\left( {AB\overline C } \right) + p\left( {AC\overline B } \right) + p\left( {ABC} \right) = \frac{1}{2} \cr & p.q.\frac{1}{2} + p.\frac{1}{2}\left( {1 - q} \right) + p.q.\frac{1}{2} = \frac{1}{2} \cr & \therefore \,pq + p - pq + pq = 1 \cr & \therefore \,p\left( {1 + q} \right) = 1 \cr} $$

$$P\left( {{E_0}} \right) = 0.45,\,\,\,P\left( {{E_1}} \right) = 0.05,\,\,\,P\left( {{E_2}} \right) = 0.50$$
For $$7$$ points there should be at least three instances of $$2$$ points and one of $$1$$ or $$2$$ points.
$$\therefore $$  the required probability
$$\eqalign{ & = P\left( {{E_2}{E_2}{E_2}{E_2}} \right) + P\left( {{E_2}{E_2}{E_2}{E_1}} \right) + P\left( {{E_2}{E_2}{E_1}{E_2}} \right) + P\left( {{E_2}{E_1}{E_2}{E_2}} \right) + P\left( {{E_1}{E_2}{E_2}{E_2}} \right) \cr & = {\left\{ {P\left( {{E_2}} \right)} \right\}^4} + 4{\left\{ {P\left( {{E_2}} \right)} \right\}^3}.P\left( {{E_1}} \right) \cr & = {\left( {0.50} \right)^4} + 4{\left( {0.50} \right)^3}.\left( {0.05} \right) \cr & = 0.0875 \cr} $$

$$\eqalign{ & {\text{We know,}} \cr & P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr & {\text{Consider}}\,;\,1 + P\left( {A \cap B} \right) - P\left( B \right) - P\left( A \right) \cr & = 1 - P\left( B \right) - P\left( A \right) + P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right) \cr & = 1 - P\left( {A \cup B} \right) \cr} $$

$$n\left( S \right) = 6 \times 6 \times 6.$$    Clearly, the sum varies from $$3$$ to $$18,$$  and among these $$4,\,9,\,16$$   are perfect squares.
The number of ways to get the sum $$4$$
$$=$$ the number of integral solutions of $${x_1} + {x_2} + {x_3} = 4$$    where $$1 \leqslant {x_1} \leqslant 6,\,1 \leqslant {x_2} \leqslant 6,\,1 \leqslant {x_3} \leqslant 6$$
$$\eqalign{ & = {\text{coefficient of }}{x^4}{\text{ in }}{\left( {x + {x^2} + ..... + {x^6}} \right)^3} \cr & = {\text{coefficient of }}x{\text{ in }}{\left( {\frac{{1 - {x^6}}}{{1 - x}}} \right)^3} \cr & = {\text{coefficient of }}x{\text{ in }}{\left( {1 - {x^6}} \right)^3}.{\left( {1 - x} \right)^{ - 3}} = {}^3{C_1} \cr} $$
Similarly, the number of ways to get the sum $$9$$
$$\eqalign{ & = {\text{coefficient of }}{x^6}{\text{ in }}{\left( {1 - {x^6}} \right)^3}.{\left( {1 - x} \right)^{ - 3}} \cr & = - 3 \times 1 + {}^8{C_6} \cr & = 28 - 3 \cr & = 25 \cr} $$
The number of ways to get the sum $$16$$
$$\eqalign{ & = {\text{coefficient of }}{x^{13}}{\text{ in }}{\left( {1 - {x^6}} \right)^3}.{\left( {1 - x} \right)^{ - 3}} \cr & = {\text{coefficient of }}{x^{13}}{\text{ in }}\left( {1 - 3{x^6} + 3{x^{12}} - {x^{18}}} \right).\left( {{}^2{C_0} + {}^3{C_1}x + {}^4{C_2}{x^2} + ....} \right) \cr & = {}^{15}{C_{13}} - 3 \times {}^9{C_7} + 3 \times {}^3{C_1} \cr & = 105 - 108 + 9 \cr & = 6 \cr & \therefore \,n\left( E \right) = 3 + 25 + 6 = 34 \cr & {\text{So, }}P\left( E \right) = \frac{{34}}{{6 \times 6 \times 6}} = \frac{{17}}{{108}}. \cr} $$

The required probability $$=$$ (probability of drawing a prime number) × (probability of drawing more than $$10$$  from the prime numbers)
$$\eqalign{ & = \frac{8}{{20}} \times \frac{4}{8} \cr & = \frac{1}{5} \cr} $$

$$X$$ = number of aces drawn
$$\eqalign{ & \therefore \,\,P\left( {X = 1} \right) + P\left( {X = 2} \right) \cr & = \left\{ {\frac{4}{{52}} \times \frac{{48}}{{52}} + \frac{{48}}{{52}} \times \frac{4}{{52}}} \right\} + \left\{ {\frac{4}{{52}} \times \frac{4}{{52}}} \right\} \cr & = \frac{{24}}{{169}} + \frac{1}{{169}} \cr & = \frac{{25}}{{169}} \cr} $$

The probability of getting a heart in one draw $$ = \frac{{13}}{{52}} = \frac{1}{4}$$
Similarly for a diamond, the probability $$ = \frac{1}{4}$$
The probability of getting a black card in one draw $$ = \frac{{26}}{{52}} = \frac{1}{2}$$
$$\therefore $$  the required probability $$ = {}^6{C_2} \times {}^4{C_2} \times {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{2}} \right)^2},$$         because in $$6$$ draws, $$2$$ draws will be for hearts, $$2$$ for diamonds and $$2$$ for black cards, and this selection can be done in $${}^6{C_2} \times {}^4{C_2} \times {}^2{C_2}$$    ways.

$$\eqalign{ & {\text{Let }}P\left( {\overline B } \right) = x.{\text{ Then }}P\left( B \right) = 1 - x \cr & \therefore \,P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right) = \frac{1}{5}\left( {1 - x} \right) \cr & {\text{But }}P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr & \Rightarrow \frac{7}{{10}} = \frac{1}{5} + \left( {1 - x} \right) - \frac{1}{5}\left( {1 - x} \right){\text{ or }}\frac{1}{2} = \frac{4}{5}\left( {1 - x} \right) \cr & \therefore \,1 - x = \frac{5}{8}\,\,\,\,\,\,\,\,\,\,\,\therefore \,x = \frac{3}{8} \cr} $$

$$p$$ = $$P$$ (correct answer), $$q$$ = $$P$$ (wrong answer)
$$ \Rightarrow \,\,p = \frac{1}{3},q = \frac{2}{3},n = 5$$
By using Binomial distribution
Required probability $$ = {\,^5}{C_4}{\left( {\frac{1}{3}} \right)^4}.\frac{2}{3} + {\,^5}{C_5}{\left( {\frac{1}{3}} \right)^5}$$
$$\eqalign{ & = 5.\frac{2}{{{3^5}}} + \frac{1}{{{3^5}}} \cr & = \frac{{11}}{{{3^5}}} \cr} $$