A father has $$3$$ children with at least one boy. The probability that he has $$2$$ boys and $$1$$ girl is :
A.
$$\frac{1}{4}$$
B.
$$\frac{1}{3}$$
C.
$$\frac{2}{3}$$
D.
none of these
Answer :
$$\frac{1}{3}$$
Solution :
Consider the following events :
$$A :$$ Father has at least one boy
$$B :$$ Father has $$2$$ boys and one girl
Then, $$A =$$ one boy and $$2$$ girls, $$2$$ boys and one girl, $$3$$ boys and no girl $$A \cap B = 2$$ boys and one girl
Now, the required probability is
$$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = \frac{1}{3}.$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$