Mutually exclusive part two

When two events are mutually exclusive then

P(A or B) = P(A) + P(B)

In our card example P(7 or a queen) = P(7) + P(queen)

P(7) = \frac{1}{13}  and P(queen) = \frac{1}{13}

P(7  or  queen) = P(7) + P(queen)

= \frac{1}{13} + \frac{1}{13} = \frac{2}{13}

We can see why this works if we look at the sample space for picking a card from a deck

Sample space for picking a card from a deck

We can see that the 7s and the queens in the sample space do not cross over. We can see that if you pick a 7 then it can’t be a queen. It is clear that there are four ways to pick a 7 and four ways to pick a queen and that there are 4 + 4 = 8 ways to pick a 7 or a queen.


7s

Queens

7s or queens

We can also see that the 7s and clubs in the sample space do cross over. There are four ways to pick a 7 and thirteen ways to pick a club but there are sixteen ways to pick a 7 or a club which is less than 4 + 13. We need to be careful not to count the 7 of clubs twice as this is a member of both events. These events are not mutually exclusive and so P(7 or club) is not equal to P(7) + P(club).


7s

Clubs

7s or clubs