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consider a bag which contains 5 balls of which 3 are red and 2 are white. Balls are randomly removed one at a time without replacement until all red red balls are drawn or until all white balls are drawn. What is the probability that last ball drawn is white color?

can someone explain how to solve this question please! The given answer is 6/10. I have made out possible cases where we draw 1)red,red,red (red=R and white =W) 2)RWRW 3)WW 4)WRW 5)RRWR All of the above are total possible cases where we stop draw all red balls or all white balls and then stop drawing out more balls. Then I have calculated the sample space of the event we need by adding up all possible outcomes from above possibilities, but i got struck there, I wasnt able to get correct sample space. When added up all of them i am getting an decimal value for sample spcae. although i guess i might have done some calculation mistake! can someone help me solve this.

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  • $\begingroup$ What have you tried so far? Please edit the question to show us what you've been able to figure out. $\endgroup$
    – Amaan M
    Jul 23, 2021 at 21:00
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    $\begingroup$ The brute force approach will work. List all 10 permutations possible. Count those which meet your criterion for stopping and then count how many end with a white ball. $\endgroup$ Jul 23, 2021 at 21:03
  • $\begingroup$ yeah i have done the brute force way but i seem to be going wrong somewhere with the calculation. it would be great if anyone gives a detailed solution! $\endgroup$
    – kafka yash
    Jul 23, 2021 at 21:12
  • $\begingroup$ Remember there are $3$ combinations that result in RWRW, because any of the three red balls can be the leftover, and even more permutations. You have to be careful when counting the sample space to be sure that each possible outcome is equally likely. It helps to arbitrarily number the balls. $\endgroup$ Jul 23, 2021 at 21:34
  • $\begingroup$ yes you are right about counting all possible permutations and i actually did include it in my calculation but the problem i went wrong and i got confused while doing that for those 5 cases and i am not sure how to proceed! i hope i can get a detailed solution. $\endgroup$
    – kafka yash
    Jul 23, 2021 at 21:45

2 Answers 2

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Here's an easy way to see the answer. Number the balls and list all $120$ possible orders. Start selecting your order by choosing the last ball to be drawn. It will be red $\frac 35$ of the time and white $\frac 25$ of the time, so the last ball to be drawn will be white with probability $\frac 35$ and it will be red with probability $\frac 25$.

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  • $\begingroup$ Ingenious use of conditions ! +1 $\endgroup$ Jul 24, 2021 at 6:37
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Method $1$:

There are $6$ situations where the last ball is white: $WW, WRW, RWW, RRWW, RWRW, WRRW$.

Hence, $P = \frac{2 \times 1}{5 \times 4} + 2(\frac{3 \times 2 \times 1}{5 \times 4 \times 3}) + 3(\frac{3 \times 2 \times 2 \times 1}{5 \times 4 \times 3 \times 2}) = \frac{6}{10}$.

$$$$

Method $2$:

There are $6$ situations where the last ball is white: $WW, WRW, RWW, RRWW, RWRW, WRRW$.

There are $4$ situations where the last ball is red: $RRR, RWRR, WRRR, RRWR$.

Hence, $P = \frac{6}{6 + 4} = \frac{6}{10}$.

$$$$

Note $1$: By computing the probabilities of each of the situations, we can see that they are all equally likely, thus validating method $2$.

Note $2$: If you use the brute force approach, try to be systematic. If you notice above, the combinations listed progressively increase in the number of balls.

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  • $\begingroup$ What makes you think that each of those situations is equally likely? $\endgroup$ Jul 23, 2021 at 22:27
  • $\begingroup$ @RobertShore Admittedly, I am wrong. I have edited my answer. $\endgroup$
    – user905694
    Jul 23, 2021 at 22:44
  • $\begingroup$ Same problem. I'm not saying they're not equally likely. I'm saying that requires proof. $\endgroup$ Jul 23, 2021 at 23:14
  • $\begingroup$ @RobertShore Hahaha, I am laughing at my own stupidity. I have edited my answer. $\endgroup$
    – user905694
    Jul 24, 2021 at 0:48

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