A box contains 24 identical balls of which 12 are white and 12 are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the $${4^{th}}$$ time on the $${7^{th}}$$ draw is
A.
$$\frac{5}{{64}}$$
B.
$$\frac{27}{{32}}$$
C.
$$\frac{5}{{32}}$$
D.
$$\frac{1}{{2}}$$
Answer :
$$\frac{5}{{32}}$$
Solution :
Prob. of a getting a white ball in a single draw
$$ = p = \frac{{12}}{{24}} = \frac{1}{2}$$
Prob. of getting a white ball $${4^{th}}$$ time in the $${7^{th}}$$ draw
= $$P$$ (getting 3 $$W$$ in 6 draws) $$ \times $$ $$P$$ (getting $$W$$ ball at $${7^{th}}$$ draw)
$$ = {\,^6}{C_3}{\left( {\frac{1}{2}} \right)^6}.\frac{1}{2} = \frac{5}{{32}}$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$