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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

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  • $\begingroup$ You can use the binomial distribution for this $\endgroup$ – DMH16 Feb 14 at 11:29
  • $\begingroup$ @DMH16 Directly Binomial (and nothing else) is wrong in this case. The $7$th draw is specified to be a white ball. $\endgroup$ – Lee David Chung Lin Feb 14 at 11:30
  • $\begingroup$ @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 $\endgroup$ – DMH16 Feb 14 at 11:32
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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 $$ { 6 \choose 3} \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^3 \frac{1}{2} = \frac{5}{32}. $$

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    $\begingroup$ Thanks so much. small correction in last line $\displaystyle \frac{5}{32}$ $\endgroup$ – jacky Feb 14 at 11:38

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