# One bag con

Given:

The bag I contains 4 yellow and 5 red balls.

Bag II contains 6 yellow and 3 red balls.

A ball is transferred from bag I to bag II and then a ball is drawn from bag II.

There are two mutually exclusive ways to draw a yellow ball from bag II –

a. A yellow ball is transferred from the bag I to bag II, and then, a yellow ball is drawn from bag II

b. A red ball is transferred from the bag I to bag II, and then, a yellow ball is drawn from bag II

Let E1 be the event that yellow ball is drawn from the bag I and E2 be the event that red ball is drawn from the bag I.

Now, we have

We also have

Let E3 denote the event that yellow ball is drawn from bag II.

Hence, we have

We also have

Using the theorem of total probability, we get

P(E3) = P(E1)P(E3|E1) + P(E2)P(E3|E2)

Thus, the probability of the drawn ball being yellow is.

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