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When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.

The multiplication rule

If A and B are two events defined on a sample space , then: P ( A B ) = P ( B ) P ( A | B ) .

This rule may also be written as: P ( A | B ) = P ( A B ) P ( B )

(The probability of A given B equals the probability of A and B divided by the probability of B .)

If A and B are independent , then P ( A | B ) = P ( A ) . Then P ( A B ) = P ( A | B ) P ( B ) becomes P ( A B ) = P ( A ) ( B ) .

One easy way to remember the multiplication rule is that the word "and" means that the event has to satisfy two conditions. For example the name drawn from the class roster is to be both a female and a sophomore. It is harder to satisfy two conditions than only one and of course when we multiply fractions the result is always smaller. This reflects the increasing difficulty satisfying two conditions.

The addition rule

If A and B are defined on a sample space, then: P ( A B ) = P ( A ) + P ( B ) - P ( A B ) .

If A and B are mutually exclusive , then P ( A B ) = 0 . Then P ( A B ) = P ( A ) + P ( B ) - P ( A B ) becomes P ( A B ) = P ( A ) + P ( B ) .

Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska

  • Klaus can only afford one vacation. The probability that he chooses A is P ( A ) = 0.6 and the probability that he chooses B is P ( B ) = 0.35.
  • P ( A B ) = 0 because Klaus can only afford to take one vacation
  • Therefore, the probability that he chooses either New Zealand or Alaska is P ( A B ) = P ( A ) + P ( B ) = 0.6 + 0.35 = 0.95 . Note that the probability that he does not choose to go anywhere on vacation must be 0.05.

Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P ( A ) = 0.65. B = the event Carlos is successful on his second attempt. P ( B ) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal | that he made the first goal is 0.90.

a. What is the probability that he makes both goals?

a. The problem is asking you to find P ( A B ) = P ( B A ) . Since P ( B | A ) = 0.90: P ( B A ) = P ( B | A ) P ( A ) = (0.90)(0.65) = 0.585

Carlos makes the first and second goals with probability 0.585.

b. What is the probability that Carlos makes either the first goal or the second goal?

b. The problem is asking you to find P ( A B ).

P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0.65 + 0.65 - 0.585 = 0.715

Carlos makes either the first goal or the second goal with probability 0.715.

c. Are A and B independent?

c. No, they are not, because P ( B A ) = 0.585.

P ( B ) P ( A ) = (0.65)(0.65) = 0.423

0.423 ≠ 0.585 = P ( B A )

So, P ( B A ) is not equal to P ( B ) P ( A ).

d. Are A and B mutually exclusive?

d. No, they are not because P ( A B ) = 0.585.

To be mutually exclusive, P ( A B ) must equal zero.

Try it

Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P ( C ) = 0.75. D = the event Helen makes the second shot. P ( D ) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?

P ( D | C ) = 0.85

P ( C D ) = P ( D C )
P ( D C ) = P ( D | C ) P ( C ) = (0.85)(0.75) = 0.6375
Helen makes the first and second free throws with probability 0.6375.

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Source:  OpenStax, Business statistics -- bsta 200 -- humber college -- version 2016reva -- draft 2016-04-04. OpenStax CNX. Apr 05, 2016 Download for free at http://legacy.cnx.org/content/col11969/1.5
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