A man draws a card from a pack of $$52$$ cards and then replaces it. After
shuffling the pack, he again draws a card. This he repeats a number of times. The probability that he will draw a heart for the first time in the third draw is :
A.
$$\frac{9}{{64}}$$
B.
$$\frac{{27}}{{64}}$$
C.
$$\frac{1}{4} \times \frac{{{}^{39}{C_2}}}{{{}^{52}{C_2}}}$$
D.
none of these
Answer :
$$\frac{9}{{64}}$$
Solution :
Let $$E = $$ the event of drawing a heart. Clearly, $$P\left( E \right) = \frac{{{}^{13}{C_1}}}{{{}^{52}{C_1}}} = \frac{1}{4}.$$
$$\therefore $$ the required probability $$ = P\left( {\overline E \,\overline E \,E} \right) = P\left( {\overline E } \right).P\left( {\overline E } \right).P\left( E \right) = \frac{3}{4}.\frac{3}{4}.\frac{1}{4} = \frac{9}{{64}}.$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$