# probability of white balls is drawn in $7$ th draw

A box contain $$24$$ identical balls, of which $$12$$ white and $$12$$ blacks. The balls are drawn at random from the box one at a time with replacement, The probability that a white ball is drawn $$4$$ th time on the seventh draw is

what i try

probability of white ball is drawn is $$\displaystyle \frac{\binom{12}{1}}{\binom{24}{1}}=\frac{12}{24}=\frac{1}{2}$$

probability of black ball is drawn $$\displaystyle 1-\frac{1}{2}=\frac{1}{2}$$

How do i solve it Help me please

• You can use the binomial distribution for this – DMH16 Feb 14 at 11:29
• @DMH16 Directly Binomial (and nothing else) is wrong in this case. The $7$th draw is specified to be a white ball. – Lee David Chung Lin Feb 14 at 11:30
• @LeeDavidChungLin Whoops I read the problem wrong, I thought the OP meant 4 white draws in a total of 7 draws, didn’t notice that the last should be a white one. Anyways, you can still use a binomial, just fixing the 7th position as a drawing a white one; it’s a trivial fix. My comment either way was not wrong. I simply stated that you need to use the binomial, which in fact, you do – DMH16 Feb 14 at 11:32

Since the successive trials are independent, and since $$P( \mbox{White} ) = P( \mbox{Black} ) = \frac{1}{2},$$ therefore the probability of $$4$$th white on the $$7$$th draw is $${ 6 \choose 3} \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^3 \frac{1}{2} = \frac{5}{32}.$$
• Thanks so much. small correction in last line $\displaystyle \frac{5}{32}$ – jacky Feb 14 at 11:38