Probability MCQ Questions & Answers in Statistics and Probability | Maths

$$n\left( S \right) = {\,^{20}}{C_4}$$
Statement - 1 :
common difference is 1; total number of cases = 17
common difference is 2; total number of cases = 14
common difference is 3; total number of cases = 11
common difference is 4; total number of cases = 8
common difference is 5; total number of cases = 5
common difference is 6; total number of cases = 2
Prob. $$ = \frac{{17 + 14 + 11 + 8 + 5 + 2}}{{^{20}{C_4}}}$$
$$ = \frac{1}{{85}}$$
Statement - 2 is false, because common difference can be 6 also.

In single throw of a dice, probability of getting 1 is = $$\frac{1}{6}$$
and prob. of not getting 1 is $$\frac{5}{6}$$
Then getting 1 in even no. of chances = getting 1 in $${2^{nd}}$$ chance or in $${4^{th}}$$ chance or in $${6^{th}}$$ chance and so on
∴ Req. Prob. $$ = \frac{5}{6} \times \frac{1}{6} + {\left( {\frac{5}{6}} \right)^3} \times \frac{1}{6} + {\left( {\frac{5}{6}} \right)^5} \times \frac{1}{6} + .....\,\infty $$
$$\eqalign{ & = \frac{5}{{36}}\left[ {\frac{1}{{1 - \frac{{25}}{{36}}}}} \right] \cr & = \frac{5}{{36}} \times \frac{{36}}{{11}} \cr & = \frac{5}{{11}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & P\left( A \right) + P\left( B \right) - P\left( A \right)P\left( B \right) = P\left( {A \cup B} \right) \cr & {\text{Comparing with }} \cr & P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) = P\left( {A \cup B} \right) \cr & {\text{we get, }}P\left( {A \cap B} \right) = P\left( A \right).P\left( B \right) \cr & \therefore \,A{\text{ and }}B{\text{ independent events}}{\text{.}} \cr} $$

Let the probability of occurrence of first event $$A$$, be $$'a'$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{., }}P\left( A \right) = a \cr & \therefore \,P\left( {{\text{not }}A} \right) = 1 - a \cr} $$
And also suppose that probability of occurrence of second event $$B,\,P\left( B \right) = b,$$
$$\eqalign{ & \therefore \,P\left( {{\text{not }}B} \right) = 1 - b \cr & {\text{Now,}} \cr & P\left( {A{\text{ and not }}B} \right) + P\left( {{\text{not }}A{\text{ and }}B} \right) = \frac{{26}}{{49}} \cr & \Rightarrow P\left( A \right) \times P\left( {{\text{not }}B} \right) + P\left( {{\text{not }}A} \right) \times P\left( B \right) = \frac{{26}}{{49}} \cr & \Rightarrow a + b - 2ab = \frac{{26}}{{49}}......\left( {\text{i}} \right) \cr & {\text{And }}P\left( {{\text{not }}A{\text{ and not }}B} \right) = \frac{{15}}{{49}} \cr & \Rightarrow P\left( {{\text{not }}A} \right) \times P\left( {{\text{not }}B} \right) = \frac{{15}}{{49}} \cr & \Rightarrow 1 - b - a + ab = \frac{{15}}{{49}} \cr & \Rightarrow a + b - ab = \frac{{34}}{{49}}......\left( {{\text{ii}}} \right) \cr & {\text{From}}\,{\text{equations }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right), \cr & a + b = \frac{{42}}{{49}}......\left( {{\text{iii}}} \right) \cr & {\text{and }}ab = \frac{8}{{49}} \cr & {\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab \cr & = \frac{{42}}{{49}} \times \frac{{42}}{{49}} - \frac{{4 \times 8}}{{49}} = \frac{{196}}{{2401}} \cr & \therefore \,a - b = \frac{{14}}{{49}}......\left( {{\text{iv}}} \right) \cr & {\text{From}}\,{\text{equations }}\left( {{\text{iii}}} \right){\text{ and }}\left( {{\text{iv}}} \right), \cr & a = \frac{4}{7},\,\,b = \frac{2}{7} \cr} $$
Hence probability of more probable of the two events $$ = \frac{4}{7}$$

We have

$$\eqalign{ & \because \,E \cap F \cap G = \phi \cr & P\left( {{E^c} \cap {F^c}/G} \right) = \frac{{P\left( {{E^c} \cap {F^c} \cap G} \right)}}{{P\left( G \right)}} \cr & = \frac{{P\left( G \right) - P\left( {E \cap G} \right) - P\left( {G \cap F} \right)}}{{P\left( G \right)}} \cr & \left[ {{\text{From ven diagram }}{E^c} \cap {F^c} \cap G = G - E \cap G - F \cap G} \right] \cr & = \frac{{P\left( G \right) - P\left( E \right)P\left( G \right) - P\left( G \right)P\left( F \right)}}{{P\left( G \right)}} \cr & = 1 - P\left( E \right) - P\left( F \right) \cr & = P\left( {{E^c}} \right) - P\left( F \right) \cr & \left[ {\because \,E,\,F,\,G{\text{ are pairwise independent}}} \right] \cr} $$

Let $$b$$ and $$g$$ represent the boy and the girl child, respectively. If a family has two children, the sample space will be
$$S = \left\{ {\left( {b,\,b} \right),\,\left( {b,\,g} \right),\,\left( {g,\,b} \right),\,\left( {g,\,g} \right)} \right\}$$
Let $$A$$ be the event that both children are girls.
Therefore, $$A = \left\{ {\left( {g,\,g} \right)} \right\}$$
$$\left( {\text{i}} \right)$$ Let $$B$$ be the event that the youngest child is a girl.
Therefore, $$B = \left\{ {\left( {b,\,g} \right),\,\left( {g,\,g} \right)} \right\}$$
$$\eqalign{ & \Rightarrow A \cap B = \left\{ {\left( {g,\,g} \right)} \right\} \cr & \therefore \,P\left( B \right) = \frac{2}{4} = \frac{1}{2}{\text{ and }}P\left( {A \cap B} \right) = \frac{1}{4} \cr} $$
The conditional probability that both are girls, given that the youngest child is a girl, is given by $$P\left( {A/B} \right).$$
$$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{{\frac{1}{4}}}{{\frac{1}{2}}} = \frac{1}{2}$$
$$\left( {{\text{ii}}} \right)$$ Let $$C$$ be the event that at least one child is a girl.
Therefore, $$C = \left\{ {\left( {b,\,g} \right),\,\left( {g,\,b} \right),\,\left( {g,\,g} \right)} \right\}$$
$$\eqalign{ & \Rightarrow A \cap C = \left\{ {g,\,g} \right\} \cr & \Rightarrow P\left( C \right) = \frac{3}{4}{\text{ and}}\,P\left( {A \cap C} \right) = \frac{1}{4} \cr} $$
The conditional probability that both are girls, given that at least one child is a girl, is given by $$P\left( {A|C} \right).$$
Therefore $$P\left( {A|C} \right) = \frac{{P\left( {A \cap C} \right)}}{{P\left( C \right)}} = \frac{{\frac{1}{4}}}{{\frac{3}{4}}} = \frac{1}{3}$$

The required probability $$=$$ (probability of the first being faulty) × (probability of the second being faulty when the first is faulty)
$$\eqalign{ & = \frac{2}{4} \times \frac{1}{3} \cr & = \frac{1}{6} \cr} $$

$$A$$ and $$B$$ will contradict each other if one speaks truth and other false . So , the required
Probability $$ = \frac{4}{{10}}\left( {1 - \frac{3}{4}} \right) + \left( {1 - \frac{4}{5}} \right)\frac{3}{4}$$
$$\eqalign{ & = \frac{4}{5} \times \frac{1}{4} + \frac{1}{5} \times \frac{3}{4} \cr & = \frac{7}{{20}} \cr} $$

$$n\left( S \right) = 2 \times 2 \times 2 \times 2 = 16,$$       because each of the four places can be filled in $$2$$ ways.
The zero determinants are \[\left| \begin{array}{l} \,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ - 1\,\,\,\, - 1 \end{array} \right|,\left| \begin{array}{l} - 1\,\,\,\,\, - 1\\ \,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right|,\left| \begin{array}{l} 1\,\,\,\,\,1\\ 1\,\,\,\,\,1 \end{array} \right|,\left| \begin{array}{l} - 1\,\,\,\,\, - 1\\ - 1\,\,\,\,\, - 1 \end{array} \right|,\left| \begin{array}{l} - 1\,\,\,\,\,1\\ - 1\,\,\,\,\,1 \end{array} \right|,\left| \begin{array}{l} 1\,\,\,\,\, - 1\\ 1\,\,\,\,\, - 1 \end{array} \right|,\left| \begin{array}{l} - 1\,\,\,\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,1\,\,\,\, - 1 \end{array} \right|,\left| \begin{array}{l} \,\,\,\,1\,\,\,\,\, - 1\\ - 1\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right|\]
$$\therefore \,\,P\left( E \right) = \frac{8}{{16}} = \frac{1}{2}.$$

Out of 6 vertices 3 can be chosen in $$^6{C_3}$$  ways.
$$\Delta $$ will be equilateral if it is $$\Delta \,ACE$$   or $$\Delta \,BDF$$   (2 ways)

∴ Required prob. $$ = \frac{2}{{^6{C_3}}} = \frac{2}{{20}} = \frac{1}{{10}}$$