There are two boxes labeled by 1 and 2 Box 1 contains 3 appl

There are two boxes labeled by 1 and 2. Box 1 contains 3 apples and 5 oranges. Box 2 contains 4 apples and 7 oranges. We flip a coin. If it is Heads, then we select one piece of fruit from Box 1; otherwise, we select one piece of fruit from Box 2. Suppose that this experiment is done, and we know that an apple was selected. What is a probability that the coin flip was Tails?

Solution

Let A be the event of selecting apple in box1 i.e., P(A)=3/8 & B be the event of selecting apple in box2 i.e., P(B)=4/11. Now by using Baye\'s theorem, the prior probability of selecting box1 is 1/2 & the prior probability of selecting box2 is 1/2 . Let C be the event of selecting an apple by flipping a coin getting head, then P(C/A)=[(1/2)*(1/3)]/[((1/2)*(1/3))+((1/2)*(1/4))]=4/7 & P(C/B)=[(1/2)*(1/4)]/[((1/2)*(1/3))+((1/2)*(1/4))]=3/7. Then P(C)=[(3/8)*(4/7)]+[(4/11)*(3/7)]=0.3701. Then P(flipping a coin getting tail) =1-P(flipping a coin by getting a heads)=1-0.3701=0.6299.