Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$n\left( S \right) = 8!$$
$$n\left( E \right) = $$   the number of arrangements of $$5$$ boys and $$3$$ girls when the $$3$$ girls are consecutive $$ = 6! \times 3!$$
$$\therefore $$  the required probability $$ = \frac{{6! \times 3!}}{{8!}} = \frac{3}{{28}}.$$

Let $${E_1}$$ be the event that the answer is guessed, $${E_2}$$ be the event that the answer is copied, $${E_3}$$ be the event that the examinee knows the answer and $$E$$ be the event that the examinee answers correctly.
Given $$P\left( {{E_1}} \right) = \frac{1}{3},\,\,P\left( {{E_2}} \right) = \frac{1}{6}$$
Assume that events $${E_1},\,{E_2}\,\,\& \,\,{E_3}$$   are exhaustive.
$$\eqalign{ & \therefore \,P\left( {{E_1}} \right) + P\left( {{E_2}} \right) + P\left( {{E_3}} \right) = 1 \cr & \therefore \,P\left( {{E_3}} \right) = 1 - P\left( {{E_1}} \right) - P\left( {{E_2}} \right) \cr & = 1 - \frac{1}{3} - \frac{1}{6} \cr & = \frac{1}{2} \cr} $$
Now, $$P\left( {\frac{E}{{{E_1}}}} \right) \equiv $$   Probability of getting correct answer by guessing $$ = \frac{1}{4}$$  (Since $$4$$ alternatives)
$$P\left( {\frac{E}{{{E_2}}}} \right) \equiv $$   Probability of answering correctly by copying $$ = \frac{1}{8}$$
and $$P\left( {\frac{E}{{{E_3}}}} \right) \equiv $$   Probability of answering correctly by knowing $$= 1$$
Clearly, $$\left( {\frac{{{E_3}}}{E}} \right)$$  is the event he knew the answer to the question given that he correctly answered it.
Using Baye’s theorem $$P\left( {\frac{{{E_3}}}{E}} \right)$$
$$\eqalign{ & = \frac{{P\left( {{E_3}} \right).P\left( {\frac{E}{{{E_3}}}} \right)}}{{P\left( {{E_1}} \right).P\left( {\frac{E}{{{E_1}}}} \right) + P\left( {{E_2}} \right).P\left( {\frac{E}{{{E_2}}}} \right) + P\left( {{E_3}} \right).P\left( {\frac{E}{{{E_3}}}} \right)}} \cr & = \frac{{\frac{1}{2} \times 1}}{{\frac{1}{3} \times \frac{1}{4} + \frac{1}{6} \times \frac{1}{8} + \frac{1}{2} \times 1}} \cr & = \frac{{24}}{{29}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & P\left( A \right) = \frac{1}{2}\,\,\,\therefore P\left( {\overline A } \right) = \frac{1}{2} \cr & P\left( B \right) = \frac{1}{3}\,\,\,\therefore P\left( {\overline B } \right) = \frac{2}{3} \cr & P\left( C \right) = \frac{1}{4}\,\,\,\therefore P\left( {\overline C } \right) = \frac{3}{4} \cr} $$
Now problem will be solved if any one of them will solve the problem,
$$\therefore \,P$$  (at least one of them solve the problem) $$ = 1 - $$  probability none of them can solve the problem or
$$\eqalign{ & P\left( {A \cup B \cup C} \right) \cr & = 1 - P\left( {\overline A } \right) \times P\left( {\overline B } \right) \times P\left( {\overline C } \right) \cr & = 1 - \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \cr & = 1 - \frac{1}{4} \cr & = \frac{3}{4} \cr} $$

Let the second success occur at the $${n^{th}}$$ trial. This means that there was exactly one success in the first $$n - 1$$  trials, so that the probability of getting the second success at the $${n^{th}}$$ trial is
$${p_n} = \left( {{}^{n - 1}{C_1}\,p{q^{n - 1 - 1}}} \right)p = \left( {n - 1} \right){p^2}{q^{n - 2}}$$
Therefore the probability of the required event is
$$\eqalign{ & {p_4} + {p_5} + {p_6} + .... = 3{p^2}{q^2} + 4{p^2}{q^3} + 5{p^2}{q^4} + 6{p^2}{q^5} + ..... \cr & = {p^2}{q^2}\left( {3 + 4q + 5{q^2} + 6{q^3} + .....} \right) \cr & = {p^2}{q^2}\left[ {3\left( {1 + q + {q^2} + {q^3} + .....} \right) + q\left( {1 + 2q + 3{q^2} + .....} \right)} \right] \cr & = {p^2}{q^2}\left[ {3{{\left( {1 - q} \right)}^{ - 1}} + q{{\left( {1 - q} \right)}^{ - 2}}} \right] \cr & = {p^2}{q^2}\left( {\frac{3}{p} + \frac{q}{{{p^2}}}} \right) \cr & = {q^2}\left( {3p + q} \right) \cr & = {q^2}\left( {2P + p + q} \right) \cr & = {q^2}\left( {2p + 1} \right) \cr & = {\left( {\frac{3}{4}} \right)^2}\left[ {2\left( {\frac{1}{4}} \right) + 1} \right] \cr & = \frac{9}{{16}}\left( {\frac{1}{2} + 1} \right) \cr & = \frac{{27}}{{32}} \cr} $$

$$n\left( S \right) = $$  the number of ways $${\left( {33} \right)^n}$$  can have a digit in the units place
$$\eqalign{ & = 4\left( {\because \,{\text{ it is }}3,\,9,\,7\,{\text{or}}\,1} \right) \cr & n\left( E \right) = 1.{\text{ So, }}P\left( E \right) = \frac{1}{4}. \cr} $$

The required event occurs if two sixes are observed in the first seven throws and a six is observed on the eighth throw. If $$p$$ is the probability that a six shows on the die, the number of throws $$n$$ is $$7$$, and $$X$$ is the number of times a six is observed, then $$X \sim B\left( {7,\,p} \right).$$
Therefore the required probability equals $$P\left( {X = 2} \right)$$   times the probability of getting a six on the eighth throw, i.e., it equals
$$\eqalign{ & = \left( {{}^7{C_2}\,{p^2}{q^5}} \right)\left( p \right) \cr & = \left( {{}^7{C_2}} \right){\left( {\frac{1}{6}} \right)^2}{\left( {\frac{5}{6}} \right)^5}\left( {\frac{1}{6}} \right) \cr & = \frac{{{}^7{C_2}\left( {{5^5}} \right)}}{{{6^8}}} \cr} $$

$$P$$ (exactly one of $$A$$ or $$B$$ occurs)
$$ = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) = \frac{1}{4}\,\,\,\,.....\left( 1 \right)$$
$$P$$ (Exactly one of $$B$$ or $$C$$ occurs)
$$ = P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) = \frac{1}{4}\,\,\,\,.....\left( 2 \right)$$
$$P$$ (Exactly one of $$C$$ or $$A$$ occurs)
$$ = P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) = \frac{1}{4}\,\,\,.....\left( 3 \right)$$
Adding (1), (2) and (3), we get
$$\eqalign{ & 2\sum {P\left( A \right) - 2\sum {P\left( {A \cap B} \right) = \frac{3}{4}} } \cr & \therefore \,\,\sum {P\left( A \right) - \sum {P\left( {A \cap B} \right) = \frac{3}{8}} } \cr & {\text{Now, }}P\left( {A \cap B \cap C} \right) = \frac{1}{{16}} \cr & \therefore \,\,P\left( {A \cup B \cup C} \right) \cr & = \sum {P\left( A \right) - \sum {P\left( {A \cap B} \right) + P\left( {A \cap B \cap C} \right)} } \cr & = \frac{3}{8} + \frac{1}{{16}} \cr & = \frac{7}{{16}} \cr} $$

$$P$$ (at least 7 pts) = $$P$$ (7pts) + $$P$$ (8 pts)
[ $$\because $$ At most 8 pts can be scored. ]
Now 7 pts can be scored by scoring 2 pts in 3 matches and 1 pt. in one match, similarly 8 pts can be scored by scoring 2 pts in each of the 4 matches.
$$\eqalign{ & \therefore \,\,{\text{Req}}{\text{. prob}}{\text{.}} = {\,^4}{C_1} \times {\left[ {P\left( {2{\text{pts}}} \right)} \right]^3}P\left( {1{\text{pts}}} \right) + {\left[ {P\left( {2{\text{pts}}} \right)} \right]^4} \cr & = 4{\left( {0.5} \right)^3} \times 0.05 + {\left( {0.50} \right)^4} \cr & = 0.0250 + 0.0625 \cr & = 0.0875 \cr} $$

The number of possible outcomes of $$2n$$  tosses is $${2^{2n}}.$$  There are $${}^n{C_r}$$ ways of getting $$r$$ heads, with $$0 \leqslant r \leqslant n,$$   in $$n$$ tosses. Therefore, the number of ways of getting $$r$$ heads in both the first $$n$$ and last $$n$$ tosses is $${\left( {{}^n{C_r}} \right)^2}.$$  Summing over all values of $$r$$, the number of favourable ways is
$${\left( {{}^n{C_0}} \right)^2} + {\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2} + ...... + {\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n},$$
So that the required probability is $$\frac{{{}^{2n}{C_n}}}{{{2^{2n}}}}.$$

Let $$X \sim B\left( {100,\,p} \right)$$    be the number of coins showing heads, and let $$q = 1 - p.$$
Then, since $$P\left( {X = 51} \right) = P\left( {X = 50} \right),$$     we have
$$\eqalign{ & {}^{100}{C_{51}}\left( {{p^{51}}} \right)\left( {{q^{49}}} \right) = {}^{100}{C_{50}}\left( {{p^{50}}} \right)\left( {{q^{50}}} \right) \cr & \Rightarrow \frac{p}{q} = \left( {\frac{{100!}}{{50!\,50!}}} \right)\left( {\frac{{51!\,49!}}{{100!}}} \right) \cr & \Rightarrow \frac{p}{{1 - p}} = \frac{{51}}{{50}} \cr & \Rightarrow 50p = 51 - 51p \cr & \Rightarrow p = \frac{{51}}{{101}} \cr} $$