Today we talked about conditional probability. There is a formula for conditional probability:
\[P(\text{A} \, | \, \text{B}) = \frac{P( \text{A and B})}{P(B)}.\]
The short-hand \(P(\text{A} \, | \, \text{B})\) means: The probability of A given B.
Row and column percentages in two-way table are typically conditional probabilities. We did a couple examples:
Suppose you draw two cards from a deck of 52 playing cards. Find \(P(\text{2nd card is Ace}\, | \, \text{First card is an Ace})\).
Our second example was taken from the textbook. It was based on data from a 1721 smallpox outbreak in Boston. Read about it here.
Suppose that 13% of students in one class got an A on the midterm exam. Of those, 47% also got an A on the final exam. For the students who didn’t get an A on the midterm, only 11% got an A on the final. Make a tree diagram for this situation and use it to find the probability that someone got an A on the midterm given that they had an A on the final exam.
A large study was conducted to determine whether mammograms are effective at detecting breast cancer. From the study, 0.8% of women aged 40-50 have breast cancer. Mammograms are 90% effective at detecting breast cancer if it is present. Mammograms are 93% accurate when breast cancer is not present. If a woman tests positive, then what is the conditional probability that she actually has breast cancer?
The key for both examples 3 and 4 is to make a tree diagram and find the probability of the given. Then the conditional probability is the probability of both A and B happening divided by the probability of the given. \[P(\text{A} \, | \, \text{B}) = \frac{P(\text{Both})}{P(\text{Given})}.\] Notice that we call event \(\text{B}\) the given.
Today we started by reviewing weighted averages.
Solution. \(0.05(100) + 0.45(73) + 0.20(4) + \frac{1}{11}(1)\).
To compute a weighted average, follow these two steps:
It is helpful to realize that weighted averages can be used to computer regular averages:
Solution. \(\frac{4}{11}(3) + \frac{4}{11}(4) +\frac{2}{11}(5) + \frac{1}{11}(1)\).
Notice that the weights can be either fractions or decimals.
There are two kinds of averages in probability theory.
As the number of trials in a random experiment increases, the sample mean \(\bar{x}\) tends to get closer and closer to the theoretical average \(\mu\).
Solution. \(\frac{18}{38}(2) + \frac{20}{38}(0) = \$0.947\).
Solution. The expected value will be less than $1,000 since car insurance companies make a profit. So they charge customers more then they give back in average reimbursements.
We finished by talking about why car insurance is a good thing to spend your money on and gambling is considered bad. The difference is risk. Car insurance reduces your risk while gambling increases your risk. Good investments are a trade off between higher expected return and higher risk. Well talk more next time about how we can measure risk mathematically.