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If there are two red, three blue and four green balls in a bag and if three balls are drawn successively without replacement then probability of getting first draw as blue ball and third draw as red ball is what?

I did $P(\text{BRR})+P(\text{BBR})+P(\text{BGR})=\frac{42}{504}$

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  • $\begingroup$ Welcome to Math.SE. What have you tried? Where are you stuck? $\endgroup$ – John Aug 19 '16 at 18:34
  • $\begingroup$ I am getting the answer as 42/504, but the options dont mention that. $\endgroup$ – user185995 Aug 19 '16 at 18:36
  • $\begingroup$ P(BRR)+P(BBR)+P(BGR)=42/504 $\endgroup$ – user185995 Aug 19 '16 at 18:38
  • $\begingroup$ Thanks. Reconfirmed the answer. $\endgroup$ – user185995 Aug 19 '16 at 18:55
  • $\begingroup$ It's generally better if you post the steps you took too and also if you type up the post too. Formatting tips here. $\endgroup$ – Em. Aug 19 '16 at 18:56
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The probability that the first draw is blue is $1/3$.

Drawing from what's left, the probability that the next two balls are red is $(2/8)(1/7) = 1/28$.

Drawing from what's left, the probability that the second ball is not red, but the third ball is, is $(6/8)(2/7) = 3/14$.

So the overall probability is

$$P = \frac{1}{3}\frac{1}{4} = \frac{1}{12}.$$

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  • $\begingroup$ How is probability of first draw as blue 1/4? Shouldn't it be 3/9? $\endgroup$ – user185995 Aug 19 '16 at 18:50
  • $\begingroup$ Dang it, I can't add. O_o Fixing now. $\endgroup$ – John Aug 19 '16 at 18:51
  • $\begingroup$ Fixed. Sorry about that! $\endgroup$ – John Aug 19 '16 at 18:53
  • $\begingroup$ Oh, and by the way, that's the same answer you got, just in lowest terms! :) $\endgroup$ – John Aug 19 '16 at 18:54

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