Math 121 - Week 1 Notes

Monday, September 14

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:

  1. Suppose you draw two cards from a deck of 52 playing cards. Find \(P(\text{2nd card is Ace}\, | \, \text{First card is an Ace})\).

  2. Our second example was taken from the textbook. It was based on data from a 1721 smallpox outbreak in Boston. Read about it here.

  3. 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.

  4. 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.

Wednesday, September 16

Today we started by reviewing weighted averages.

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:

  1. Multiply each outcome by its weight.
  2. Add the results.

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.

Averages in Probability

There are two kinds of averages in probability theory.

  1. Sample mean (\(\bar{x}\)) is the average of the observed outcomes in a random experiment.
  2. Theoretical average (\(\mu\)) is the weighted average of the outcomes using the probabilities as the weights. The theoretical average is also known as the expected value.

The Law of Large Numbers

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. \(\frac{18}{38}(2) + \frac{20}{38}(0) = \$0.947\).

  1. If you pay $1,000 each year for car insurance, will the expected value of the money you get back from the car insurance company be greater or less than $1,000?

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.