Warm-up – for my history buffs… A general can plan a campaign to fight one major battle or three small battles. He believes that he has probability 0.6 of winning the large battle and probability of 0.8 winning each of the small battles. Victories or defeats in the small battles are independent. The general must win either the large battle or all three small battles to win the campaign. Which strategy should he choose?

Wrapping Up Chapter 6 Tree Diagrams

Tree Diagrams Suppose that 2% of a clinic’s patients are known to be HIV+. A blood test is developed that is positive in 98% of patients with HIV, but is also positive in 3% of patients without HIV. Find P(positive test). Find P(positive test ∩ HIV+) If a person who is chosen at random from the clinic’s patients is given the test and it comes out positive, what is the probability that the person actually has HIV?

Decisions, decisions! Many probability problems involve making a decision. A tree diagram can help us organize the information.

Dialysis or Transplant? Lynn has to decide between dialysis or a kidney transplant. Here are the facts: 52% of dialysis patients survive for 3 years. After 1 month, 96% of kidney transplants succeed. 3% fail to function, and 1% die. Patients who return to dialysis still have a 52% chance of surviving 3 years. Of the successful transplants, 82% continue to function for 3 years. 8% must return to dialysis, of whom 70% survive to the 3 year mark. 10% of the successful transplants die without returning to dialysis.

Example The probability of rain today is .3. Also, 40% of all rainy days are followed by rainy days and 20% of all days without rain are followed by rainy days. The following tree diagram represents the weather for today and tomorrow. Complete a tree diagram to represent this situation. What is the probability that it rains on both days? What is the probability that it rains on one of the two days? What is the probability that it does not rain on either day?

Testing for Independence Remember the general rule for multiplication: P(A∩B) = P(A)*P(B|A) Also remember the multiplication rule for independent events: P(A∩B) = P(A)*P(B) if A and B are independent.

There are two ways to test for independence: P(A∩B) = P(A)*P(B|A) If A and B are independent, then P(A∩B) = P(A)*P(B) Therefore, by substitution, if A and B are independent, then P(B|A) = P(B)

Here is the table of Central High’s student population. Male Female Total Seniors 156 144 300 Juniors 168 172 340 Sophomores 196 164 360 520 480 a. What is the probability of selecting a male? b. What is the probability of selecting a male from the sophomore class? c. Use your answers from parts a and b to determine whether the events “selecting a male” and “selecting a sophomore” are independent.

Homework Chapter 5 # 63, 71, 73, 85, 88

HW answers 63. a) .2428 b) .5983 71. a) .41 b).6341 73. b < b/t < t < t/b 85. .1423 88. a) diagram b) .6387