A cricket club has $$15$$ members, of whom only $$5$$ can bowl. If the names of $$15$$ members are put into a box and $$11$$ are drawn at random, then the probability of getting an eleven containing at least $$3$$ bowlers is :
A.
$$\frac{7}{{13}}$$
B.
$$\frac{6}{{13}}$$
C.
$$\frac{{11}}{{15}}$$
D.
$$\frac{{12}}{{13}}$$
Answer :
$$\frac{{12}}{{13}}$$
Solution :
The total number of ways of choosing $$11$$ players out of $$15$$ is $${}^{15}{C_{11}}.$$ A team of $$11$$ players containing at least $$3$$ bowlers can be chosen in the following mutually exclusive ways :
$$\left( {{\bf{I}}} \right)$$ Three bowlers out of $$5$$ bowlers and $$8$$ other players out of the remaining $$10$$ players.
$$\left( {{\bf{II}}} \right)$$ Four bowlers out of $$5$$ bowlers and $$7$$ other players out of the remaining $$10$$ players.
$$\left( {{\bf{III}}} \right)$$ Five bowlers out of $$5$$ bowlers and $$6$$ other players out of the remaining $$10$$ players.
So, required probability
$$\eqalign{
& = P\left( {\bf{I}} \right) + P\left( {{\bf{II}}} \right) + P\left( {{\bf{III}}} \right) \cr
& = \frac{{{}^5{C_3} \times {}^{10}{C_8}}}{{{}^{15}{C_{11}}}} + \frac{{{}^5{C_4} \times {}^{10}{C_7}}}{{{}^{15}{C_{11}}}} + \frac{{{}^5{C_5} \times {}^{10}{C_6}}}{{{}^{15}{C_{11}}}} \cr
& = \frac{{12}}{{13}} \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$