4.3 Two basic rules of probability  (Page 19/4)

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 $\text{P(A)}=0.6$ and the probability that he chooses $B$ is $\text{P(B)}=0.35$ .
• $\text{P(A and B)}=0$ because Klaus can only afford to take one vacation
• Therefore, the probability that he chooses either New Zealand or Alaska is $\text{P(A OR B)}=\text{P(A)}+\text{P(B)}=0.6+0.35=0.95$ . 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. $\text{P(A)}=0.65$ . $B$ = the event Carlos is successful on his second attempt. $\text{P(B)}=0.65$ . Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice 4 times a week. Thirty of the intermediate swimmers practice 4 times a week. Ten of the novice swimmers practice 4 times a week. Suppose one member of the swim team is randomly chosen. Answer the questions (Verify the answers):

Studies show that, if she lives to be 90, about 1 woman in 7 (approximately 14.3%) will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let $B$ = woman develops breast cancer and let $N$ = tests negative. Suppose one woman is selected at random.