I have a question pertaining to the binomial distribution and conditional probability. I managed to figure out the answer intuitively but have trouble applying and getting the same answer using the conditional probability formula.
3 machines A, B and C produce 25%, 35% and 40% respectively of the golf balls manufactured by a factory. These balls are either yellow or white. Of the balls produced by A and B, 20% and 30% respectively are yellow. It is known that the probability of picking a yellow ball is 0.355. If 3 balls are picked randomly, find the probability that at least 1 is yellow given that all the balls picked are from machine A.
The answer I figured out using intuition was 0.488.
How do I apply the conditional probability formula to this case? $ P(A|B)=\frac{P(A\cap B)}{P(B)}$
Thank you in advance.
You want $\mathsf P(Y_1\cup Y_2\cup Y_3\mid A_1\cap A_2\cap A_3)$, the probability of picking at least one yellow ball among three balls, given that all three among those balls are from factory $A$.
Since $20\%$ of balls, from factory $A$, are yellow, this is just $1-(1-0.20)^3$. As your intuition informed you.
$$\mathsf P(Y_1\cup Y_2\cup Y_3\mid A_1\cap A_2\cap A_3)=0.488$$
You do not.
Because you can not. The formula allows you to calculate one of the conditional, joint, and marginal probabilities when you know the other two probabilities. You only know the marginal probability, $\mathsf P(A_1\cap A_2\cap A_3)=0.25^3$, but do not know the joint probability, so cannot use the formula to calculate the conditional probability.
So you must evaluate it by other means, and you have correctly done that.
Now that you have done so, you may use the conditional probability formula to calculate the joint probability:$$\begin{align}\mathsf P((Y_1\cup Y_2\cup Y_3)\cap (A_1\cap A_2\cap A_3)) &= 0.488\times (0.25)^3\\&=0.007625\end{align}$$