Question

A box contains 24 identical balls, of which 12 are black and 12 are white. The balls are drawn at random from the box one at a time with replacement. The probability that white ball is drawn for the fourth time on seventh draw is,
A). $\dfrac{5}{{64}}$
B). $\dfrac{{27}}{{32}}$
C). $\dfrac{5}{{32}}$
D). $\dfrac{1}{2}$

Answer 1

This question as it appears to be very simple and it is simple. Only we need to catch the hint that it is based on Bernauli’s principle. In which there is an exact probability of success and failure. Like here the event that a white ball is to be drawn has the success attempt and failure attempt well defined. So let's take help of Bernauli’s principle.

Complete step-by-step solution:
Given that, a box contains 24 identical balls, of which 12 are black and 12 are white.
Probability of drawing a white ball successfully is $\dfrac{{12}}{{24}} = \dfrac{1}{2}$
We know that, the probability of successful event is
$P\left( {x = r} \right) = \sum {{}^n{C_r}} {\left( p \right)^r}{\left( q \right)^{n - r}}$
Now we know that the first 3 events are definitely true. So we can take x=r=3
And we know that n=6 because the trial is going to be for the seventh attempt.
$P\left( {x = 3} \right) = \sum {{}^6{C_3}} {\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{6 - 3}}$
Taking the values,
$P\left( {x = 3} \right) = \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^3}$
We can write the factorials as and since the denominators of both the fractions has base 2 we can add the power,

$P\left( {x = 3} \right) = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 6}} \times \dfrac{1}{{{2^6}}}$
$P\left( {x = 3} \right) = 20 \times \dfrac{1}{{{2^6}}}$
At the last attempt we are sure that the ball so drawn is white. So we can say that the probability is $\dfrac{1}{2}$
$P(E) = 20 \times \dfrac{1}{{{2^6}}} \times \dfrac{1}{2}$
On solving we get,
$P(E) = \dfrac{5}{{{2^5}}}$
Taking fifth power of 2 we get,
$P(E) = \dfrac{5}{{32}}$
This is the correct answer for the solution above.
Thus option C is correct.

Note: Note that the balls are drawn at random but with replacement. Also note that we chose the principle for the first three attempts because we are sure that for the fourth attempt the ball is definitely going to be white. So I took that probability for the final attempt.