1. ## probabilty

a bag contains 10 black and 16 white balls.22 balls are drawn one by one till 4 white balls remain in the bag.what is the probability the 22th ball drawn is black. all balls are distinct.

a bag contains 10 black and 16 white balls.22 balls are drawn one by one till 4 white balls remain in the bag.what is the probability the 22th ball drawn is black. all balls are distinct.
Use Bayes theorem: You want

$P(b_{22}=b|\mbox{exactly 12 of first 22 are white})=\dfrac{P(\mbox{exactly 12 of first 22 are white}|b_{22}=b)P(b_{22}=b)}{P(\mbox{exactly 12 of first 22 are white})}$

CB

$\text{A bag contains 10 black and 16 white balls.}$
$\text{22 balls are drawn one by one untll 4 white balls remain in the bag.}$
$\text{What is the probability the 22th ball drawn is black.}$
$\text{All balls are distinct.}$ . Is this relevant?

I have a back-door approach to this problem.
I may be totally wrong, but I can't shake my reasoning.

. . . . . . . . . $\overbrace{\circ\;\bullet\;\bullet\;\circ\;\bullet \;\hdots\;\square}^{\text{10 black, 12 white}}\;|\;\circ\;\circ\;\circ\;\circ$
. . . . . . . . . . . . . . . . . . . . $\Uparrow$
What is the probability that this is black?

Isn't it . $\dfrac{10}{22} \:=\:\dfrac{5}{11}$ ?

4. will someone explain soroban sir's answer

will someone explain soroban sir's answer
It is the same thing as: Four white balls are drawn from the bag. Now what is the probability that when you draw one more ball it is black (which is the ratio of the number of black balls in the bag 10, to the total number of balls in the bag 22).

CB