In detail, a bag has 8 white, $3$ red, $2$ blue balls. One by one balls are drawn from the bag. If the first two draws result in white balls, find the probability that the third ball drawn would be white when the balls are drawn without replacement.
I think if the first two draws result in white balls then we are left with $6$ white balls and the probability of drawing a third white ball is $6 \over 11$. Is this line of thinking valid or am I missing something else?
Best Answer
You're entirely correct. Given that you've drawn two white balls, you've got the following configuration in your bag: 6 white balls, 5 non-white balls. The probability of drawing a third white ball is $\frac{6}{11}$.