Probability MCQ Questions & Answers in Statistics and Probability | Maths

We have
$$\eqalign{ & P\left( {x \geqslant 1} \right) \geqslant \frac{9}{{10}} \cr & \Rightarrow \,\,1 - P\left( {x = 0} \right) \geqslant \frac{9}{{10}} \cr & \Rightarrow \,\,1 - {\,^n}{C_0}{\left( {\frac{1}{4}} \right)^0}{\left( {\frac{3}{4}} \right)^n} \geqslant \frac{9}{{10}} \cr & \Rightarrow \,\,1 - \frac{9}{{10}} \geqslant {\left( {\frac{3}{4}} \right)^n} \cr & \Rightarrow \,\,{\left( {\frac{3}{4}} \right)^n} \leqslant \left( {\frac{1}{{10}}} \right) \cr} $$
Taking log to the base $${\frac{3}{4}},$$ on both sides, we get
$$\eqalign{ & n{\log _{\frac{3}{4}}}\left( {\frac{3}{4}} \right) \geqslant {\log _{\frac{3}{4}}}\left( {\frac{1}{{10}}} \right) \cr & \Rightarrow \,\,n \geqslant { - \log _{\frac{3}{4}}}10 \cr & = \frac{{ - {{\log }_{10}}10}}{{{{\log }_{10}}\left( {\frac{3}{4}} \right)}} \cr & = \frac{{ - 1}}{{{{\log }_{10}}3 - {{\log }_{10}}4}} \cr & \Rightarrow \,\,n \geqslant \frac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}} \cr} $$

Maximum sum of numbers appearing on four dice together $$ = 24$$
$$\therefore $$  Required probability $$ = 0$$

$$\eqalign{ & P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right) = 0.20 \cr & {\text{Also }}P\left( {A' \cap B} \right) = P\left( B \right) - P\left( {A \cap B} \right) = 0.15 \cr & \Rightarrow P\left( A \right) + P\left( B \right) - 2P\left( {A - B} \right) = 0.35 \cr & {\text{Now, }}P\left( {A' \cap B'} \right) = 1 - P\left( {A \cup B} \right) \cr & \Rightarrow 0.1 = 1 - P\left( A \right) - P\left( B \right) + P\left( {A \cap B} \right) \cr & \Rightarrow P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) = 0.9 \cr & \Rightarrow P\left( {A \cap B} \right) = 0.9 - 0.35 \cr & \Rightarrow P\left( {A \cap B} \right) = 0.55{\text{ and}} \cr & P\left( A \right) = 0.75,\,P\left( B \right) = 0.70 \cr & {\text{Now, }}P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{{0.55}}{{0.70}} = \frac{{11}}{{14}} \cr} $$

Probability that at least one shot hits the plane
$$\eqalign{ & = 1 - P\left( {{\text{none of the shot hits the plane}}} \right) \cr & = 1 - 0.6 \times 0.7 \times 0.8 \times 0.9 \cr & = 1 - 0.3024 \cr & = 0.6976 \cr} $$

Desired probability $$=$$ probability of getting $$3$$ sixes in first $$9$$ throws $$ \times $$ getting six in the $$10$$  throw
$$ = {}^9{C_3}{\left( {\frac{1}{6}} \right)^3}{\left( {\frac{5}{6}} \right)^6} \times \frac{1}{6} = \frac{{84 \times {5^6}}}{{{6^{10}}}}$$

Probability of getting a composite number is $$\frac{2}{6} = \frac{1}{3}$$
$$\eqalign{ & {\text{Probability that }}A{\text{ will win the game is}} \cr & \left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + .... \cr & = \frac{{\frac{1}{3}}}{{1 - \frac{8}{{27}}}} \cr & = \left( {\frac{1}{3} \times \frac{{27}}{{19}}} \right) \cr & = \frac{9}{{19}} \cr & {\text{Probability that }}B{\text{ will win the game is}} \cr & \left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + .... \cr & = \frac{{\frac{2}{9}}}{{1 - \frac{8}{{27}}}} \cr & = \left( {\frac{2}{9} \times \frac{{27}}{{19}}} \right) \cr & = \frac{6}{{19}} \cr & {\text{Probability that }}C{\text{ will win the game is}} \cr & \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + .... \cr & = \frac{{\frac{4}{{27}}}}{{1 - \frac{8}{{27}}}} \cr & = \left( {\frac{4}{{27}} \times \frac{{27}}{{19}}} \right) \cr & = \frac{4}{{19}} \cr & {\text{So required ratio is}}\,\,9:6:4 \cr} $$

Here, $$P\left( {A \cup B} \right) = \frac{3}{5}$$    and $$P\left( {A \cap B} \right) = \frac{1}{5}.$$    So, from the addition theorem,
$$\eqalign{ & \frac{3}{5} = P\left( A \right) + P\left( B \right) - \frac{1}{5} \cr & {\text{or }}\frac{4}{5} = 1 - P\left( {A'} \right) + 1 - P\left( {B'} \right) \cr & \therefore \,P\left( {A'} \right) + P\left( {B'} \right) = 2 - \frac{4}{5} = \frac{6}{5} \cr} $$

Numbers divisible by $$6$$ are $$6,\,12,\,18,\,......,\,90.$$
Numbers divisible by $$8$$ are $$8,\,16,\,24,\,......,\,88.$$
Now, total number divisible by $$6 = 15$$
and total number divisible by $$8 = 11$$
Now, the number divisible by both $$6$$ and $$8$$ are $$24,\,48,\,72.$$
So, total number divisible by both $$6$$ and $$8 = 3$$
$$\therefore $$  Probability ( number divisible by $$6$$ or $$8$$ )
$$ = \frac{{15 + 11 - 3}}{{90}} = \frac{{23}}{{90}}$$

Let $$P\left( A \right) = $$   probability that a randomly selected student likes chocolate $$ = 0.5$$
Let $$P\left( B \right) = $$   probability that a randomly selected student likes cake $$ = 0.3$$
Then $$P\left( {A \cap B} \right) = 0.1$$
Now we have to find $$P\left( {\frac{A}{B}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{{0.1}}{{0.3}} = \frac{1}{3}$$

There are $$30$$  days in April.
$$n\left( S \right) = $$   the number of ways in which $$10$$  persons can have birthdays in the month of April
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 30 \times 30 \times .....$$     to $$10$$  times $$ = {30^{10}}$$  (because each person can have birthday in $$30$$  ways).
$$n\left( E \right) = n\left( S \right) - $$    the number of ways in which $$10$$  persons can have different birthdays
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {30} \right)^{10}} - {}^{30}{C_{10}} \cr & \therefore \,P\left( E \right) = \frac{{{{\left( {30} \right)}^{10}} - {}^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}. \cr} $$