Probability MCQ Questions & Answers in Statistics and Probability | Maths

Given : Probability of aeroplane $$I$$, scoring a target correctly i.e., $$P\left( I \right) = 0.3$$   probability of scoring a target correctly by aeroplane $$II$$, i.e. $$P\left( {II} \right) = 0.2$$
$$\eqalign{ & \therefore \,P\left( {\overline I } \right) = 1 - 0.3 = 0.7 \cr & \therefore \,{\text{The required probability}} \cr & = P\left( {\overline I \cap II} \right) \cr & = P\left( {\overline I } \right).P\left( {II} \right) \cr & = 0.7 \times 0.2 \cr & = 0.14 \cr} $$

If a no. is to be divisible by both 2 and 3. It should be divisible by their L.C.M.
∴ L.C.M. of (2 and 3) = 6
∴ Numbers are = 6, 12,18 . . . . 96.
Total numbers are = 16
∴ Probability $$ = \frac{{^{16}{C_3}}}{{^{100}{C_3}}} = \frac{4}{{1155}}$$

The required probability $$ = 1 - $$   probability of each receiving at least one
$$ = 1 - \frac{{n\left( E \right)}}{{n\left( S \right)}}.$$
Now, the number of integral solutions of $${x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = 10$$        such that $${x_1} \geqslant 1,\,{x_2} \geqslant 1,\,......,\,{x_6} \geqslant 1$$       gives $$n\left( E \right)$$  and the number of integral solutions of $${x_1} + {x_2} + ...... + {x_5} + {x_6} = 10$$       such that $${x_1} \geqslant 0,\,{x_2} \geqslant 0,\,......,\,{x_6} \geqslant 0$$       gives $$n\left( S \right).$$
$$\therefore $$  the required probability $$ = 1 - \frac{{{}^{10 - 1}{C_{6 - 1}}}}{{{}^{10 + 6 - 1}{C_{6 - 1}}}} = 1 - \frac{{{}^9{C_5}}}{{{}^{15}{C_5}}} = \frac{{137}}{{143}}.$$

$$n\left( S \right) = $$   total number of sections of two diagonals $$ = {}^{{}^6{C_2} - 6}{C_2} = {}^9{C_2} = \frac{{9 \times 8}}{2} = 36.$$
$$n\left( E \right) = $$   the number of selections of two diagonals which intersect at an interior point
           $$ = $$ the number of selections of four vertices $$ = {}^6{C_4} = 15$$
$$\therefore $$  the required probability $$ = \frac{{15}}{{36}} = \frac{5}{{12}}.$$

Let $$A$$ be the event that $${1^{st}}$$ man will be alive till $$75$$  years and $$B$$ be the event that $${2^{nd}}$$ man will be alive till $$70$$  years then $$P\left( A \right) = \frac{5}{{16}}$$   and $$P\left( B \right) = \frac{3}{7}$$   then $$P\left( {A'} \right) = \frac{{11}}{{16}}$$   and $$P\left( {B'} \right) = \frac{4}{7}.$$
Probability that none of them will be alive $$35$$  years hence is $$P\left( {A'} \right) \times P\left( {B'} \right) = \frac{{11}}{{16}} \times \frac{4}{7} = \frac{{11}}{{28}}$$
Then required probability is $$1 - \frac{{11}}{{28}} = \frac{{17}}{{28}}$$

The number of ways of arranging $$50$$  books$$ = {}^{50}{P_{50}} = 50!.$$
The number of ways of choosing places for the five volume dictionary is $${}^{50}{C_5}$$  and the number of ways of arranging the remaining $$45$$  books $$ = {}^{45}{P_{45}} = \left( {45} \right)!.$$
Thus the number of favourable ways is $$\left( {{}^{50}{C_5}} \right)\left( {45!} \right).$$
Hence the probability of the required event
$$\eqalign{ & = \frac{{\left( {{}^{50}{C_5}} \right)\left( {45!} \right)}}{{50!}} \cr & = \left( {\frac{{50!}}{{5!\,45!}}} \right)\left( {\frac{{45!}}{{50!}}} \right) \cr & = \frac{1}{{5!}} \cr & = \frac{1}{{120}} \cr} $$

Total number of non negative solutions of $$x + y + z = 10$$    are $$^{12}{C_2} = 66$$   (using $$^{n + r - 1}{C_{r - 1}}$$  )
If $$z$$ is even then there can be following cases :
$$z = 0$$
⇒ No. of ways of solving $$x + y = 10$$
⇒ $$^{11}{C_1}$$
$$z = 2$$ ⇒ No. of ways of solving $$x + y = 8$$
⇒ $$^{9}{C_1}$$
$$z = 4$$
⇒ No. of ways of solving $$x + y = 6$$
⇒ $$^{7}{C_1}$$
$$z = 6$$
⇒ No. of ways of solving $$x + y = 4$$
⇒ $$^{5}{C_1}$$
$$z = 8$$
⇒ No. of ways of solving $$x + y = 2$$
⇒ $$^{3}{C_1}$$
$$z = 10$$
⇒ No. of ways of solving $$x + y = 0$$
⇒ 1
∴ Total ways when $$z$$ is even = 11 + 9 + 7 + 5 + 3 + 1 = 36
∴ Required probability $$ = \frac{{36}}{{66}} = \frac{6}{{11}}$$

Given that $$P$$ (India wins) $$= p =$$   $$\frac{1}{2}$$
∴ $$P$$ (India loses) $$= p’ =$$   $$\frac{1}{2}$$
Out of 5 matches india’s second win occurs at third test
⇒ India wins third test and simultaneously it has won one match from first two and lost the other.
∴ Required prob. = $$P (LWW) + P (WLW)$$
$$ = {\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^3} = \frac{1}{4}$$

$$\eqalign{ & {\text{Events }}A{\text{ and }}B{\text{ are mutually exclusive}}{\text{.}} \cr & {\text{Hence,}}\,P\left( {A \cap B} \right) = \phi = 0 \cr & \therefore \,P\left( {A \cup B} \right) + P\left( A \right) + P\left( B \right)......\left( 1 \right) \cr & P\left( A \right) = 0.2\,\,\,\,\,\,\,\,\left[ {{\text{given}}} \right] \cr & P\left( B \right) = P\left( {\overline A \cap B} \right) + P\left( {A \cap B} \right) \cr & \Rightarrow P\left( B \right) = P\left( {\overline A \cap B} \right)\,\,\,\,\,\,\,\,\,\,\left[ {\because P\left( {A \cap B} \right) = 0} \right] \cr & \Rightarrow P\left( B \right) = 0.3 \cr & P\left( {A \cup B} \right) = 0.2 + 0.3 = 0.5 \cr & P\left( {A\left| {\left( {A \cup B} \right)} \right.} \right) = \frac{{P\left( A \right)}}{{P\left( {A \cup B} \right)}} = \frac{{0.2}}{{0.5}} \cr & \Rightarrow P\left( {A\left| {\left( {A \cup B} \right)} \right.} \right) = \frac{2}{5} \cr} $$

Let $$A$$ denote the event that the person has TB.
Let $$B$$ denote the event that the person has not TB.
Let $$E$$ denote the event that the person is diagnosed to have TB.
$$\eqalign{ & \therefore \,P\left( A \right) = \frac{1}{{1000}},\,P\left( B \right) = \frac{{999}}{{1000}} \cr & P\left( {\frac{E}{A}} \right) = 0.99,\,P\left( {\frac{E}{B}} \right) = 0.001 \cr & {\text{The required probability is given by}} \cr & P\left( {\frac{A}{E}} \right) = \frac{{P\left( A \right) \times \left( {\frac{E}{A}} \right)}}{{P\left( A \right) \times P\left( {\frac{E}{A}} \right) + P\left( B \right) \times \left( {\frac{E}{B}} \right)}} \cr & = \frac{{\frac{1}{{1000}} \times 0.99}}{{\frac{1}{{1000}} \times 0.99 + \frac{{999}}{{1000}} \times 0.001}} \cr & = \frac{{110}}{{221}} \cr} $$