# Two approaches to the same problem in probability

#### Yankel

##### New Member
Hello all

I have this question:

Two basketball players shoot the ball. The probability of the first one to succeed is 0.8, and the probability of the second is 0.6. What is the probability that only one of them succeeded ?

I solved this problem in two ways.

Firstly, I realized that the events

A - The first one succeeded
B - The second one succeeded

Are independent. Therefore, (~)A and (~)B are also independent (~ is not). I realized that A and not B, and B and not A, are disjoint, so I just used multiplication and addition to get 0.44.

Then I tried another way. I have calculated the probability of A and B using the independence assumption to get 0.48. Then I used venn diagram to find that A and not B is 0.32 and B and not A is 0.12, giving 0.44.

I could also use union of A and B, and subtract the intersection.

I wanted to ask you if there is an explanation of why the two approach work simultaneously. I mean, if the events are independent, how is the relation between the events looks like, can I draw it?

Thank you