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.how to solve this question by Baye's theorem
I have looked at this question several times.
I think that there has been no replies because none of us knows what it means.
If I had to say what is asked here it is:
What is the probability that after taking twenty-two balls out of the bag there are four white balls left and the last ball is black.
But I really doubt that is a correct reading. That has nothing to do with Baye's theorem.
Please, please try to enlighten us as to what it means.
I agree, it's rather involved, but if we *have to* use Bayes' theorem, we'd get:
$\displaystyle P(22B | last4W) = {P(last4W | 22B) P(22B) \over P(last4W | 22B) P(22B) + P(last4W | 22W) P(22W)}$
This can be calculated in a fairly straight forward manner.
The real straight forward method (that can be used for verification) would be:
P(22B | last4W) = P(first is B | total 10 B and 22 W balls)
Hello, ayushdadhwal!
I replied to this problem elsewhere.
I'm still not sure if my interpretation is correct.
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.
How to solve this question by Bayes' theorem?
To me, the situation looks like this:
. . $\displaystyle \overbrace{\bullet\:\circ\:\bullet\:\bullet\: \hdots\: \circ\:\boxdot}^{\text{10 black, 12 white}} \:| \: \circ\:\circ\:\circ\;\circ$
= . . . . . . . . . . . $\displaystyle \uparrow$
. . . . . . . . . . $\displaystyle ^{\text{Is this black?}}$