Mathematics 375 – Probability and Statistics 1

Solutions For Whole Class Discussion Problems – The “Event-Composition Method” September 16, 2011

Discussion Questions

A) Given: P (A) = .2, P (B) = .3, P (A ∪ B) = .4. Determine:

1) P (\overline{A} ∪ \overline{B}). Solution: By the additive law, we have

P (\overline{A} ∪ \overline{B}) = P (\overline{A}) + P (\overline{B}) − P (\overline{A} ∩ \overline{B}).

Since P (A) = .2, P (\overline{A}) = 1 − .2 = .8. Similarly, P (\overline{B}) = 1 − .3 = .7. Finally, \overline{A} ∩ \overline{B} = \overline{A ∪ B} by one of the DeMorgan Laws from set theory. So P (\overline{A} ∩ \overline{B}) = 1 − .4 = .6. Putting everything together,

P (\overline{A} ∪ \overline{B}) = .8 + .7 − .6 = .9

2) P (\overline{A}|B). Solution: By the definition of conditional probability,

P (\overline{A}|B) = \frac{P (\overline{A} ∩ B)}{P (B)}

We know P (B) = .3 from the given information. In the course of solving part A, we saw P (\overline{A} ∩ \overline{B}) = .6 and P (\overline{A}) = .8. Since

\overline{A} = (\overline{A} ∩ B) ∪ (\overline{A} ∩ \overline{B}),

and the two intersections in the parentheses don’t intersect each other, we have

.8 = P (\overline{A} ∩ B) + .6, hence P (\overline{A} ∩ B) = .2

Then

P (\overline{A}|B) = \frac{.2}{.3} ≐ .67.

B) The circles marked 1,2,3,4 in the diagrams on the sheet given in class represent electrical relays that operate independently and function properly with probability p = .85. Which design yields the higher probability that current will flow when the relays are activated?

Solution: Let 1, 2, 3, 4 represent the events that the corresponding relays function correctly. Layout I allows the current to flow if either 1 and 2 function correctly, or 3 and 4 do, or both sets of two do. So the probability we want is P ((1 ∩ 2) ∪ (3 ∩ 4)). By the additive law and independence, this is

\begin{aligned}P (1 ∩ 2) + P (3 ∩ 4) − P (1 ∩ 2 ∩ 3 ∩ 4) & = (.85)^{2} + (.85)^{2} − (.85)^{4} \\ & ≐ .923. \end{aligned}

Similarly, Layout II allows the current to flow if any one of the 4 relays functions correctly. The current does not flow only if all four relays fail, so we can compute the probability we want as

1 − P (\overline{1} ∩ \overline{2} ∩ \overline{3} ∩ \overline{4}) = 1 − (.15)^{4} ≐ .9995.

As you should have seen intuitively, Layout II gives a significantly higher probability.

C) An accident victim will die unless he receives type A+ blood (strictly) before 8 minutes elapse. He will be saved if he does get the blood transfusion. Potential donors and a reusable blood typing kit are available, but it takes 2 minutes to determine each possible donor’s blood type, and only 40% of them have type A+ blood. What is the probability that the victim will be saved if only one donor’s blood can be typed at a time? (Follow-up: Are you making an assumption in your solution?)

Solution: There are just three chances for an A+ donor to be found since each typing takes 2 minutes and the blood has to be given to the victim strictly before 8 minutes have passed. The chance of finding a good donor on the first test is .4. When that test fails, there is a (.6)(.4) chance of success on the second test. Similarly, when the first two tests fail, there is a (.6)^{2}(.4) chance of success on the last test before time runs out. The total probability of finding a donor and saving the victim is

(.4) + (.4)(.6) + (.4)(.6)^{2} = .784

More formally, say F is the event that a match for the blood is found on or before the third try. Then

F = F_{1} ∪ F_{2} ∪ F_{3}

where F_{i} = first match found on try i, for i = 1, 2, 3. Then F_{i} ∩ F_{j} = ∅ if i ≠ j, so

P (F) = P (F_{1}) + P (F_{2}) + P (F_{3}).

As before P (F_{1}) = .4. But then F_{2} = match found on the second person but not on the first. So P (F_{2}) is the probability that the first person tested does not have type A+ blood, and the second person does have type A+ blood. The value (.4)(.6) is OK here (only) as long as the two people’s blood types are independent. Similarly for the (.4)(.6)^{2} in the last term.