Three numbers are chosen at random without replacement from {1, 2, 3, . . . . . 8}. The probability that their minimum is 3, given that their maximum is 6, is:
A.
$$\frac{3}{8}$$
B.
$$\frac{1}{5}$$
C.
$$\frac{1}{4}$$
D.
$$\frac{2}{5}$$
Answer :
$$\frac{1}{5}$$
Solution :
Given sample space = {1, 2, 3, . . . . . ,8}
Let Event
$$A$$ : Maximum of three numbers is 6.
$$B$$ : Minimum of three numbers is 3.
This is the case of conditional probability
We have to find $$P$$ (minimum) is 3 when it is given that $$P$$ (maximum) is 6.
$$\eqalign{
& \therefore \,\,P\left( {\frac{B}{A}} \right) = \frac{{P\left( {B \cap A} \right)}}{{P\left( A \right)}} \cr
& = \frac{{\frac{{^2{C_1}}}{{^8{C_3}}}}}{{\frac{{^5{C_2}}}{{^8{C_3}}}}} \cr
& = \frac{{^2{C_1}}}{{^5{C_2}}} \cr
& = \frac{2}{{10}} \cr
& = \frac{1}{5} \cr} $$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$