Hello, kingman!

I can't imagine how to draw a Venn diagram for this problem.

First, probability that P is correct: .John has 5 items named .

Each item is placed in a separate box and sealed.

He then addresses the boxes at random, one each to

1) Find the probability that the item is in the correct box

and item is in the wrong box. (3/20)

Then probability that Q is incorrect: .

Therefore: .

Bayes' Theorem: . .[1]2) Find the probability that the item is in the correct box,

given that the item is in the wrong box. (3/16)

We found the numerator in part (1): .

The denominator is: .

Substitute into [1]: .

This is a bit trickier . . .3) Find the probability that both and are in the wrong boxes. (13/20)

There are two scenarios:

. . (1) goes to

. . (2) doesgo tonot

(1) goes to

. . .We have: .

. . The other four items can haveorderany

. . and will be in the wrong box.

. . .Hence: .

(2) doesgo tonot

. . . . . . . . . . . .

. . .

. . . can go to any of the last three addresses: .

. . .Then can go to three of the remaining four addresses: .

. . . Hence: .

Therefore: .