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