Probability MCQ Questions & Answers in Statistics and Probability | Maths
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181.
Two numbers are selected randomly from the set $$S = \left\{ {1,2,3,4,5,6} \right\}$$ without replacement one by one. The probability that minimum of the two numbers is less than 4 is
A
$$\frac{1}{{15}}$$
B
$$\frac{14}{{15}}$$
C
$$\frac{1}{{5}}$$
D
$$\frac{4}{{5}}$$
Answer :
$$\frac{4}{{5}}$$
The minimum of two numbers will be less than 4 if at least one of the numbers is less than 4.
∴ $$P$$ (at least one no. is < 4),
= $$1 - P$$ (both the no’s are $$ \geqslant $$ 4)
$$\eqalign{
& = 1 - \frac{3}{6} \times \frac{2}{5} \cr
& = 1 - \frac{6}{{30}} \cr
& = 1 - \frac{1}{5} \cr
& = \frac{4}{5} \cr} $$
182.
If $$C$$ and $$D$$ are two events such that $$C \subset D$$ and $$P\left( D \right) \ne 0,$$ then the correct statement among the following is
A
$$P\left( {\frac{C}{D}} \right) \geqslant P\left( C \right)$$
B
$$P\left( {\frac{C}{D}} \right) < P\left( C \right)$$
C
$$P\left( {\frac{C}{D}} \right) = \frac{{P\left( D \right)}}{{P\left( C \right)}}$$
D
$$P\left( {\frac{C}{D}} \right) = P\left( C \right)$$
Answer :
$$P\left( {\frac{C}{D}} \right) \geqslant P\left( C \right)$$
In this case, $$P\left( {\frac{C}{D}} \right) = \frac{{P\left( {C \cap D} \right)}}{{P\left( D \right)}} = \frac{{P\left( C \right)}}{{P\left( D \right)}}$$
Where, $$0 \leqslant P\left( D \right) \leqslant 1,$$ : hence $$P\left( {\frac{C}{D}} \right) \geqslant P\left( C \right)$$
183.
For two events $$A$$ and $$B$$ it is given that $$P\left( A \right) = P\left( {\frac{A}{B}} \right) = \frac{1}{4}{\text{ and }}P\left( {\frac{B}{A}} \right) = \frac{1}{2}.$$
Then :
A
$$A$$ and $$B$$ are mutually exclusive events
B
$$A$$ and $$B$$ are dependent events
C
$$P\overline {\left( {\frac{A}{B}} \right)} = \frac{3}{4}$$
We have, $$P\left( A \right) = P\left( {\frac{A}{B}} \right) = \frac{1}{4}$$
This shows that $$A$$ and $$B$$ are independent events.
$$\eqalign{
& {\text{So, }}P\left( B \right) = P\left( {\frac{B}{A}} \right) = \frac{1}{2} \cr
& {\text{Now, }}P\left( {\frac{A}{B}} \right) = \frac{1}{4} \cr
& \Rightarrow P\overline {\left( {\frac{A}{B}} \right)} = 1 - \frac{1}{4} = \frac{3}{4} \cr} $$
184.
The probability that a particular day in the month of July is a rainy day is $$\frac{3}{4}$$. Two person whose credibility are $$\frac{4}{5}$$ and $$\frac{2}{3}$$, respectively, claim that $${15^{th}}$$ July was a rainy day. The probability that it was really a rainy day is :
A
$$\frac{{12}}{{13}}$$
B
$$\frac{{11}}{{12}}$$
C
$$\frac{{24}}{{25}}$$
D
$$\frac{{29}}{{30}}$$
Answer :
$$\frac{{24}}{{25}}$$
Let events
$$A :$$ Event that first man speaks truth
$$B :$$ Event that second man speaks truth
$$R :$$ Day is rainy
$$\therefore \,P\left( A \right) = \frac{4}{5},\,P\left( B \right) = \frac{2}{3},\,P\left( R \right) = \frac{3}{4}$$
$$\therefore $$ Required probability
$$\eqalign{
& = \frac{{P\left( {A \cap B} \right).P\left( R \right)}}{{P\left( {A \cap B} \right).P\left( R \right) + P\left( {A' \cap B'} \right).P\left( {R'} \right)}} \cr
& = \frac{{\frac{4}{5} \times \frac{2}{3} \times \frac{3}{4}}}{{\frac{4}{5} \times \frac{2}{3} \times \frac{3}{4} + \frac{1}{5} \times \frac{1}{3} \times \frac{1}{4}}} \cr
& = \frac{{24}}{{25}} \cr} $$
185.
A boy is throwing stones at a target. The probability of hitting the target at any trial is $$\frac{1}{2}$$. The probability of hitting the target $${5^{th}}$$ time at the $${10^{th}}$$ throw is :
A
$$\frac{5}{{{2^{10}}}}$$
B
$$\frac{{63}}{{{2^9}}}$$
C
$$\frac{{{}^{10}{C_5}}}{{{2^{10}}}}$$
D
none of these
Answer :
$$\frac{{63}}{{{2^9}}}$$
The probability of hitting the target $${5^{th}}$$ time at the $${10^{th}}$$ throw $$= P$$ (the probability of hitting the target $$4$$ times in the first $$9$$ throws) $$ \times $$ (the probability of hitting the target at the $${10^{th}}$$ throw)
$$\eqalign{
& = \left[ {{}^9{C_4}{{\left( {\frac{1}{2}} \right)}^4}{{\left( {\frac{1}{2}} \right)}^5}} \right]\left( {\frac{1}{2}} \right) \cr
& = \frac{{9!}}{{4!\,5!}} \times {\left( {\frac{1}{2}} \right)^{10}} \cr
& = \frac{{63}}{{{2^9}}} \cr} $$
186.
Three numbers are chosen at random without replacement from the set $$A = \left\{ {x|1 \leqslant x \leqslant 10,\,x\, \in \,N} \right\}.$$ The probability that the minimum of the chosen numbers is $$3$$ and maximum is $$7,$$ is :
A
$$\frac{1}{{12}}$$
B
$$\frac{1}{{15}}$$
C
$$\frac{1}{{40}}$$
D
none of these
Answer :
$$\frac{1}{{40}}$$
$$n\left( S \right) = {}^{10}{C_3}$$ and $$n\left( E \right) = {}^{10}{C_1},$$ because on selecting $$3,\,7$$ and we have to select one from $$4,\,5$$ and $$6.$$
$$\therefore \,P\left( E \right) = \frac{{{}^3{C_1}}}{{{}^{10}{C_3}}} = \frac{1}{{40}}.$$
187.
Two dice are thrown n times in succession. The probability of obtaining a double-six at least once is :
The probability of getting a double -six in one throw $$ = \frac{1}{{36}}$$
The probability of not getting a double -six in one throw $$ = 1 - \frac{1}{{36}} = \frac{{35}}{{36}}$$
So the probability of not getting a double -six in $$n$$ throw $$ = {\left( {\frac{{35}}{{36}}} \right)^n}$$
$$\therefore $$ Probability of obtaining a double-six at least once $$ = 1 - {\left( {\frac{{35}}{{36}}} \right)^n}$$
188.
Let $$A, B, C$$ be three mutually independent events. Consider the two statements $${S_1}$$ and $${S_2}$$
$${S_1}$$ : $$A$$ and $$B$$ $$ \cup $$ $$C$$ are independent
$${S_2}$$ : $$A$$ and $$B$$ $$ \cap $$ $$C$$ are independent
Then,
A
Both $${S_1}$$ and $${S_2}$$ are true
B
Only $${S_1}$$ is true
C
Only $${S_2}$$ is true
D
Neither $${S_1}$$ nor $${S_2}$$ is true
Answer :
Both $${S_1}$$ and $${S_2}$$ are true
$$\eqalign{
& P\left[ {A \cap \left( {B \cup C} \right)} \right] = P\left[ {\left( {A \cap B} \right) \cup \left( {A \cap C} \right)} \right] \cr
& = P\left( {A \cap B} \right) + P\left( {A \cap C} \right) - P\left( {A \cap B \cap C} \right) \cr
& = P\left( A \right)P\left( B \right) + P\left( A \right)P\left( C \right) - P\left( A \right)P\left( B \right)P\left( C \right) \cr
& = P\left( A \right)\left[ {P\left( B \right) + P\left( C \right) - P\left( {B \cap C} \right)} \right] = P\left( A \right)P\left( {B \cup C} \right) \cr
& \therefore \,\,{S_1}\,{\text{is true}}{\text{.}} \cr
& P\left( {A \cap \left( {B \cap C} \right)} \right) = P\left( A \right)P\left( B \right)P\left( C \right) = P\left( A \right)P\left( {B \cap C} \right) \cr
& \therefore \,\,{S_2}\,{\text{is also true}}{\text{.}} \cr} $$
189.
A die is thrown. Let $$A$$ be the event that the number obtained is greater than 3. Let $$B$$ be the event that the number obtained is less than 5. Then $$P\left( {A \cup B} \right)$$ is
A
$$\frac{3}{5}$$
B
0
C
1
D
$$\frac{2}{5}$$
Answer :
1
$$A$$ $$ \equiv $$ number is greater than 3
$$ \Rightarrow \,\,P\left( A \right) = \frac{3}{6} = \frac{1}{2}$$
$$B$$ $$ \equiv $$ number is less than 5
$$ \Rightarrow \,\,P\left( B \right) = \frac{4}{6} = \frac{2}{3}$$
$$A \cap B \equiv $$ number is greater than 3 but less than 5.
$$\eqalign{
& \Rightarrow \,\,P\left( {A \cap B} \right) = \frac{1}{6} \cr
& \therefore P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr
& = \frac{1}{2} + \frac{2}{3} - \frac{1}{6} \cr
& = \frac{{3 + 4 - 1}}{6} \cr
& = 1 \cr} $$
190.
Let $$A$$ and $$B$$ be two events such that $$P\left( {\overline {A \cup B} } \right) = \frac{1}{6},P\left( {\overline {A \cap B} } \right) = \frac{1}{4}$$ and $$P\left( {\overline A } \right) = \frac{1}{4},$$ where $$\overline A $$ stands for the complement of the event $$A.$$ Then the events $$A$$ and $$B$$ are
A
independent but not equally likely.
B
independent and equally likely.
C
mutually exclusive and independent.
D
equally likely but not independent.
Answer :
independent but not equally likely.
Given
$$\eqalign{
& P\left( {\overline {A \cup B} } \right) = \frac{1}{6} \cr
& \Rightarrow \,\,P\left( {A \cup B} \right) = 1 - \frac{1}{6} = \frac{5}{6} \cr} $$
$$\eqalign{
& P\left( {\overline A } \right) = \frac{1}{4} \cr
& \Rightarrow \,\,P\left( A \right) = 1 - \frac{1}{4} = \frac{3}{4} \cr} $$
We know
$$\eqalign{
& P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr
& \Rightarrow \,\,\frac{5}{6} = \frac{3}{4} + P\left( B \right) - \frac{1}{4}\,\,\left( {\because \,P\left( {A \cap B} \right) = \frac{1}{4}} \right) \cr
& \Rightarrow \,\,P\left( B \right) = \frac{1}{3} \cr} $$
$$\because \,\,P\left( A \right) \ne P\left( B \right)$$ so they are not equally likely.
Also $$P\left( A \right) \times P\left( B \right) = \frac{3}{4} \times \frac{1}{3}$$
$$ = \frac{1}{4} = P\left( {A \cap B} \right)$$
So $$A$$ & $$B$$ are independent.