Question

Suppose P(C) = 98%; P(D) = 88%; and P(C or D) = 33%.

Find P(C and D).
Can this really be a probability? If not, explain why not.

Answer 1

`P(C\ and\ D) = 153\%`

Explanation:

Given

`P(C) = 98\%`

`P(D) = 88\%`

`P(C\ or\ D) = 33\%.`

Required

`P(C\ and\ D)`

In probability, the relationship between the given parameters and the required parameter is:

`P(C\ and\ D) = P(C) + P(D) - P(C\ or\ D)`

Substitute values in the above formula

`P(C\ and\ D) = 98\% + 88\% - 33\%`

`P(C\ and\ D) = 153\%`

No, it can't be a probability

Because
`P(C\ and\ D) > 100\%`