Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$\frac{{1 + 4p}}{4},\,\frac{{1 - p}}{2},\,\frac{{1 - 2p}}{2}$$       are probabilities of the three mutually exclusive events, then
$$\eqalign{ & 0 \leqslant \frac{{1 + 4p}}{4} \leqslant 1,\,0 \leqslant \frac{{1 - p}}{2} \leqslant 1,\,0 \leqslant \frac{{1 - 2p}}{2} \leqslant 1 \cr & {\text{and }}0 \leqslant \frac{{1 + 4p}}{4} + \frac{{1 - p}}{2} + \frac{{1 - 2p}}{2} \leqslant 1 \cr & \therefore \, - \frac{1}{4} \leqslant p \leqslant \frac{3}{4},\, - 1 \leqslant p \leqslant 1,\, - \frac{1}{2} \leqslant p \leqslant \frac{1}{2},\,\frac{1}{2} \leqslant p \leqslant \frac{5}{2} \cr & \therefore \,\frac{1}{2} \leqslant p \leqslant \frac{1}{2} \cr} $$
[The intersection of above four intervals]
$$\therefore \,p = \frac{1}{2}$$

$$n\left( S \right) = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4}$$
$$n\left( E \right) = n\left( S \right) - $$   the number of ways in which one boy gets both the pens
$$ = n\left( S \right) - {}^{10}{C_2} \times {}^8{C_4} \times {}^4{C_4} \times \left( {3!} \right)$$
$$\therefore \,P\left( E \right) = 1 - \frac{{{}^{10}{C_2} \times {}^8{C_4} \times {}^4{C_4} \times \left( {3!} \right)}}{{{}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4}}} = 1 - \frac{6}{{11}} = \frac{5}{{11}}.$$

$$\eqalign{ & {\text{As given,}} \cr & P\left( A \right) = 0.6,\,P\left( B \right) = 0.3{\text{ and }}\,P\left( {A \cap B} \right) = 0.2 \cr & P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr & = 0.6 + 0.3 - 0.2 \cr & = 0.9 - 0.2 \cr & = 0.7 \cr & {\text{So, }}P\left( {{\text{neither in }}A{\text{ nor in }}B} \right) = 1 - P\left( {A \cup B} \right) \cr & = 1 - 0.7 \cr & = 0.3 \cr} $$

$$n\left( S \right) = 100 \times 100 \times 100$$
We know that $${\left( {2n + 1} \right)^2} + {\left( {2{n^2} + 2n} \right)^2} = {\left( {2{n^2} + 2n + 1} \right)^2}$$         for all $$n\, \in \,N.$$
$$\therefore $$  for $$n = 1,\,2,\,3,\,4,\,5,\,6,$$    we get lengths of the three sides of a right-angled triangle whose longest side $$ \leqslant 100.$$
For example, when $$n = 1,$$  sides are $$3,\,4,\,5;$$  when $$n = 2,$$  sides are $$5,\,12,\,13$$   and so on.
The number of selections of $$3,\,4,\,5$$  from the three cards by taking one from each is $$3!.$$
$$\eqalign{ & \therefore \,n\left( E \right) = 6\left( {3!} \right). \cr & {\text{Hence,}}\,P\left( E \right) = \frac{{6\left( {3!} \right)}}{{100 \times 100 \times 100}} = \frac{1}{{100}}{\left( {\frac{3}{{50}}} \right)^2}. \cr} $$

Total number of ways of arranging the balls $$ = \frac{{10!}}{{3!\,7!}} = 120$$
Favourable cases, $$x{B_1}y{B_2}z{B_3}t$$
If $$x,\,y,\,z$$   and $$t$$ be the number of white balls to be places as shown above then $$0 \leqslant x,\,t \leqslant 5,\,1 \leqslant y,\,z \leqslant 6$$
$$\therefore $$  Number of favorable cases
$$\eqalign{ & = {\text{coefficient of }}{x^7}{\text{ in}}{\left( {1 + x + {x^2} + ..... + {x^5}} \right)^2}{\left( {x + {x^2} + ..... + {x^6}} \right)^2} \cr & = {\text{coefficient of }}{x^7}{\text{ in }}{x^2}{\left( {1 + x + {x^2} + ..... + {x^5}} \right)^4} \cr & = {\text{coefficient of }}{x^5}{\text{ in}}{\left( {\frac{{1 - {x^6}}}{{1 - x}}} \right)^4} \cr & = {\text{coefficient of }}{x^5}{\text{ in}}{\left( {1 - x} \right)^{ - 4}} \cr & = {}^8{C_5} \cr & = 56 \cr} $$
$$\therefore $$  Desired probability $$ = \frac{{56}}{{120}} = \frac{7}{{15}}$$

$$\eqalign{ & S = \left\{ {HHH,\,HHT,\,HTH,\,HTT,\,THH,\,THT,\,TTH,\,TTT} \right\} \cr & E = \left\{ {HHH,\,HHT,\,HTH,\,THH} \right\} \cr & F = \left\{ {HHH,\,HHT,\,HTH,\,HTT} \right\} \cr & E \cap F = \left\{ {HHH,\,HHT,\,HTH} \right\} \cr & n\left( {E \cap F} \right) = 3,\,n\left( F \right) = 4 \cr & \therefore \,{\text{Required probability }} = P\left( {\frac{E}{F}} \right) = \frac{{n\left( {E \cap F} \right)}}{{n\left( F \right)}} = \frac{3}{4} \cr} $$

According to Poission distribution, prob. of getting $$k$$ successes is
$$\eqalign{ & P\left( {x = k} \right) = {e^{ - \lambda }}\frac{{{\lambda ^k}}}{{k!}}P\left( {x \geqslant 2} \right) \cr & = 1 - P\left( {x = 0} \right) - P\left( {x = 1} \right) \cr & = 1 - {e^{ - \lambda }} - {e^{ - \lambda }}\left( {\frac{\lambda }{{1!}}} \right) \cr & = 1 - \frac{3}{{{e^2}}}. \cr} $$

$$n = 7$$
Prob. of getting any no. out 1, 2, 3, . . . . 9 is $$p = \frac{9}{{15}}$$
$$\therefore \,\,q = \frac{6}{5}$$
$$P\left( {x = 7} \right) = \,{\,^7}{C_7}{p^7}{q^0}$$     [Binomial distribution]
$$\eqalign{ & = {\left( {\frac{9}{{15}}} \right)^7} \cr & = {\left( {\frac{3}{5}} \right)^7} \cr} $$

The min. face value is not less than 2 and max. face value is not greater than 5 if we get any of the numbers 2, 3, 4, 5, while total possible out comes are 1, 2, 3, 4, 5 and 6.
∴ In one thrown of die, prob. of getting any no. Out of 2, 3, 4 and 5 is $$ = \frac{4}{6} = \frac{2}{3}$$
If the die is rolled four times, then all these events being independent, the required prob. $${\left( {\frac{2}{3}} \right)^4} = \frac{{16}}{{81}}$$

Probability of getting a blue ball at any draw $$ = p = \frac{{10}}{{20}} = \frac{1}{2}$$
$$P$$ [getting a blue ball $${4^{th}}$$ time in $${7^{th}}$$ draw]$$ = P$$  [getting $$3$$ blue balls in $$6$$ draws] $$ \times P$$  [a blue ball in the $${7^{th}}$$ draw].
$$\eqalign{ & = {}^6{C_3}{\left( {\frac{1}{2}} \right)^3}{\left( {\frac{1}{2}} \right)^3}.\frac{1}{2} \cr & = \frac{{6 \times 5 \times 4}}{{1 \times 2 \times 3}}{\left( {\frac{1}{2}} \right)^7} \cr & = 20 \times \frac{1}{{32 \times 4}} \cr & = \frac{5}{{32}} \cr} $$