How about this.

One bad contains 4 white balls and 6 black balls. Another bag contains 8 white balls and 2 black balls. A coin tossed to select a bad, then a ball is randomly selected from that bag.

a. What is the probability that a white ball will be drawn?
b. Suppose a white ball is drawn. What is the probability that it came from the First Bag?

Thanks

THE ANS : A>3/5
B>1/3

But how do you get those?

http://www.jiskha.com/display.cgi?id=1217983092

Use the ideas in Reiny's post; adjust for the numbers you're now posting.

To calculate the probabilities, we can use basic principles of probability. Let's break down the problem step by step:

a. What is the probability that a white ball will be drawn?

To find the probability of drawing a white ball, we need to consider two cases: selecting the First Bag and selecting the Second Bag.

Case 1: Selecting the First Bag:
The probability of selecting the First Bag is 1/2 since there are two bags (First Bag and Second Bag), and we toss a fair coin to determine which bag we choose.

Given that we select the First Bag, there are 4 white balls and 6 black balls. The probability of drawing a white ball from the First Bag is 4/(4+6) = 4/10 = 2/5.

Case 2: Selecting the Second Bag:
Similarly, the probability of selecting the Second Bag is also 1/2 since there are two bags.

Given that we select the Second Bag, there are 8 white balls and 2 black balls. The probability of drawing a white ball from the Second Bag is 8/(8+2) = 8/10 = 4/5.

Now, we calculate the overall probability of drawing a white ball by considering both cases:
Probability of selecting the First Bag * Probability of drawing a white ball from the First Bag +
Probability of selecting the Second Bag * Probability of drawing a white ball from the Second Bag

(1/2) * (2/5) + (1/2) * (4/5) = 1/5 + 2/5 = 3/5

Therefore, the probability of drawing a white ball is 3/5.

b. Suppose a white ball is drawn. What is the probability that it came from the First Bag?

To find the probability that the white ball came from the First Bag given that it was drawn, we can use Bayes' theorem:

Probability (White ball from First Bag | White ball drawn) =
(Probability of drawing a white ball from the First Bag * Probability of selecting the First Bag) /
Probability of drawing a white ball

We already calculated the probability of drawing a white ball from the First Bag as 2/5, the probability of selecting the First Bag as 1/2, and the probability of drawing a white ball as 3/5.

(2/5) * (1/2) / (3/5) = (1/5) / (3/5) = 1/3

Therefore, the probability that the white ball came from the First Bag given that it was drawn is 1/3.