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A bag contains 10 white balls and 3 black drawn out without replacement. Calculate the probability that all 3 black balls are drawn out by the seventh attempt.

Probability of drawing the third black ball at the seventh attempt.

$$P_1 = \frac 1 7$$

Probability of of drawing two black balls and 4 white ones in the first 6 attempts is equal to favorable draws which can be calculates by arranging the balls in a straight line $$N_2=(4W,2B)= \frac{6!}{4!2!}$$

We can calculate all possible cases using this so $p_2$ will be
$$P_2=\frac{n(4W,2B)}{n(6W)+n(5W,1B)+n(4W,2B)+n(3W,3B)}$$

$$P_2=\frac{\frac{6!}{4!2!}}{\frac{6!}{6!0!}+\frac{6!}{5!2!}+\frac{6!}{4!2!}+\frac{6!}{3!3!}}$$

$$P_2=\frac{15}{42}$$

Total probability will be $P_1 P_2 = \frac{15}{294}$ However the answer is given as $\frac{15}{286}$ Where did I ho wrong

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  • $\begingroup$ I get ${35\over286}.$ Are you certain about the given answer? $\endgroup$ – saulspatz Mar 31 at 18:25
  • $\begingroup$ @saulspatz Their solution is $$\frac {3 C 2 \cdot 10C4}{13C6} \frac {1}{ 7}$$ That's it nothing more. I think they've assumed even the identical colored balls to be distinct and proceeded with that. Seems correct to me. $\endgroup$ – user659291 Mar 31 at 18:41
  • $\begingroup$ I can't read what you've written. I'll post an answer. $\endgroup$ – saulspatz Mar 31 at 18:43
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    $\begingroup$ Well, apparently they are taking "by the seventh attempt" to mean "on the seventh attempt". $\endgroup$ – saulspatz Mar 31 at 18:53

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