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A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability the that last ball drawn is black equals to

a) $1/100\quad $ b) $1/60\quad $ c) $3/5\quad $ d) $2/3\quad $

My try:

total red balls=40 ,

total black balls=60

The probability of drawing last black ball=total black balls/(total balls)
$$=\frac{60}{40+60}=60/100=3/5$$

I guessed this answer which is correct but i don't think that my procedure is correct. I think there should be a correct solution to such problems of probability. Please give correct solution to this problem.
My book suggests that answer must be 3/5

thanks

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  • $\begingroup$ I think the intuitive solution is fine but if you're doing it for an assignment then I would say something like there are ${100 \choose 60}$ ways of picking that combination of reds and blacks. Then there are ${99 \choose 59}$ ways of choosing reds and blacks such that a black is left. I think the ratio should give you the desired fraction. $\endgroup$ – stuart stevenson Feb 7 '18 at 15:15
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    $\begingroup$ Each draw has a probability of $\frac {60}{60+40}=.6$ chance of being black. There is no special significance to being the "last". $\endgroup$ – lulu Feb 7 '18 at 15:21
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If you want to complicate matters, you could compute

$\text{(ways of placing 59 black balls in 99)/(ways of placing 100 black balls in 60)}= \dfrac{\binom{99}{59}}{\binom{100}{60}}$

but the book's suggestion is good. Imagine all the balls to be randomly placed in a line, and ask

  • what is the probability that the first ball, say, is black ? $\frac35$

  • what is the probability that the ninth ball, say, is black ? $\frac35\;\;$

  • what is the probability that the $i^{th}$ ball is black ? $\frac35\;\;$ isn't it ?

  • so what is the probability that the last ball is black ?

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