Conditional Probability: Bags and Marbles

Question

Bag A has $3$ white and $2$ black marbles. Bag B has $4$ white and $3$ black marbles.

Suppose we draw a marble at random from Bag A and put it in Bag B. After doing this, we draw a marble at random from Bag B, which turns out to be white. Given this information, what is the probability that the marble we moved from Bag A to Bag B is white?

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This problem is different from other conditional probability problems. It has a changing variable. I'm stuck on how to approach this problem. Could someone pelase walk me step by step through this problem? Thanks!

Answer 1

The law of total probability helps here. If the transferred marble was white, which happens $\frac35$ of the time, the probability that the marble drawn from bag B is white is $\frac58$. If the transferred marble was black ($\frac25$ chance), that probability is $\frac12$.

The probability that the drawn marble is white and the transferred marble is white is $\frac58×\frac35=\frac38$. The probability that the drawn marble is white but the transferred marble is black is $\frac12×\frac25=\frac15$. Therefore the probability the transferred marble was white given that the drawn marble is white is $$\frac{\dfrac38}{\dfrac38+\dfrac15}=\frac{15}{23}$$

Answer 2

Put it another way.   Bag A has three white and two black marbles of which one is drawn and placed in the second bag.   Bag B then has that marble, four white, and three black marbles, from which one is drawn.

Let $W_A$ be the event of drawing a white marble from bag A, and $W_B$ be that of drawing white from bag B.   Clearly $\mathsf P(W_A)=3/5, \mathsf P(W_A^\complement)=2/5$.

Now, when given that the marble drawn from bag A is white, bag B contains five white, and three black marbles.   However, when given that the marble drawn from bag A is black, bag B contains four white, and four black marbles.

Can you now find $\mathsf P(W_B\mid W_A)$ and $\mathsf P(W_B\mid W_A^\complement)$ ?

Then use Bayes' Rule to find $\mathsf P(W_A\mid W_B)$.