If in a bag there are 6 red balls and 10 green balls and 3 balls are drawn at random then what is the probability that the first 2 balls are green given that the third ball is red (without replacing them)? Please have a look at what I did.. And I am getting answer as $42.8$% (Correct me if wrong)
2026-03-29 13:27:23.1774790843
What is the probability that the first 2 balls are green given that the third ball is red (without replacing them)?
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Given that the third is red, we may as well consider the first two draws from a population of $5$ red and $10$ green. They are both green with probability $$\frac {10}{15}\times \frac 9{14}=\frac 37=.\overline {428571}$$ confirming your result.
Your method looks correct but needlessly complex. Once you know that the third draw is red, we have $5$ reds and $10$ greens left to distribute amongst $15$ slots. Each ball is equally likely to wind up in any of the available slots, hence the calculation given above.