I am a big advocate of using tree diagrams to solve problems involving conditional probability. However, some of these problems are easier to do if we reconstruct a frequency table from the probabilities in the problem.
Problem A doctor is studying the relationship between blood pressure and heartbeat abnormalities in her patients. She tests a random sample of her patients and notes the blood pressure (high, low, normal) and their heartbeats (regular or irregular). She finds that
When we carry out probability problems, the first task is to identify the events in the problem. Often we can identify a few key events and then use compliments, “and”, or “or” . This can simplify and help us to represent the information pictorially.
The examples below illustrate this idea.
It easy to confuse conditional probability with probability of an intersection of two event. They are related! The probability of an event A given that event B has occurred is
The vertical bar | means “given” and the event after it is the event that has already occurred.
Let’s look at some data to determine how to find several different probabilities including conditional probability.
Problem Mammograms are typically used to screen women for breast cancer. Like most medical tests, they are not perfect. Some women who do not have breast cancer have a positive mammogram. This means that they do not have cancer, but the test indicates that the do. Other women test negative on the mammogram, but do have breast cancer. A test of 10,000 women who had a mammogram gave the following results.
Assume that these data apply to all women. Now let’s define some events:
+: a woman has a positive mammogram
-: a woman has a negative mammogram
C: a woman has breast cancer
C’: a woman does not have breast cancer.
We will use these events to answer the questions below.
a. What is the probability a woman has breast cancer?
Solution In terms of our events, we are looking to calculate P (C ). To do this, we need to find the number of women with breast cancer and divide it by the number of women in the survey,
b. What is the probability that a woman has a positive mammogram?
Solution In terms of our events, we are looking to calculate P (+ ). To do this, we need to find the number of women with a positive mammogram and divide it by the number of women in the survey,
c, What is the probability that a woman has a negative mammogram and does not have breast cancer?
Solution Now things get a little more complicated. We are now interested in women with a negative mammogram and who do not have breast cancer. From the table, these are the women who are in the negative mammogram row and in the do not have cancer column, 9208. In terms of events, these are women in the event – and C’ (similarly Counting those women compared to the total number of women gives
d. If a woman has a negative mammogram, what is the probability that she does not have breast cancer?
Solution In this part, we know a woman has had a negative mammogram. Of those women, we want to know what portion does not have breast cancer. Since we know something in advance, this is a conditional probability problem. We need to calculate the probability that a woman does not have cancer, given that the woman had a negative mammogram or P (C ’ | -).
To calculate this probability, we need to take into the account the fact that we know the woman had a negative mammogram. Based on the table, we know that 9217 women had a negative mammogram. Of these women, 9208 did not have cancer. This means that
Notice that we can also think of this symbolically as
This is the same formula as
but with C’ instead of A and – instead of B.
Expected value is straightforward as long as you know all of the values of an experiment. If the values associated with each outcome are V1, V2, …, Vn and P1, P2, …, Pn are the corresponding probabilities, then the expected value is
E = P1·V1 + P2·V2 + … + Pn·Vn
This says that we need to multiply each value times the corresponding probability and then add up all of those products.
Let’s look at one example that can cause a little confusion.
Problem Assume that you have $10,000 to invest in stocks. The likelihoods of how much the stocks will change are given in the table below.
What would be the expected gain or loss (in dollars)?
Solution There is a tendency to look at the table above and to interpret the first column as the probabilities and the second column as the corresponding values. This is partially true. The first column is certainly the probabilities. For the second column to be the values, they have to be in dollars. Instead they are percentages.
To fix this problem, we need to figure out what a 6% gain is in dollars. If we are investing $10,000, a 6% gain is 6% of $10,000 or
.06 ·10,000 = 600 dollars
Let’s now update the table to have probabilities and values.
Now multiply the probabilities and add the results:
E = 0.5 · 600 + 0.3 · 0 + 0.2 · -200 = 260
This means that if you make this investment over and over, you can expect to gain $260 each time you make the investment.
If E and F are events, we can compute the probability of the union of these events using
In an applied problem, you might see the word “or” used in place of the union ∪ symbol or the word “and” used in place of the intersection symbol ∩. In this case we can write out this fu=ormula as
P(E or F) = P(E) + P(F) – P(E and F)
If we know any three of the four probabilities in the formula, we can solve for the fourth probability. This formula can also be used if the probabilities involve compliments. In that case, we can utilize
to help compute the probability. The examples below illustrates how we can use these formulas.
Problem Your bicycle has a noise coming from the back wheel. The probability that the rear hub needs to be adjusted is 0.4. The probability that the brakes need adjusting is 0.55. The probability that the rear hub needs to be adjusted and the brakes need to be adjusted is 0.10.
a. What is the probability that the rear hub needs adjusting or the rear brakes need adjusting?
Solution Let’s start by defining the events in this problem.
R: the rear hub needs adjusting
B: the brakes need adjusting
Examining the question, we notice that it is looking for the two events with “or” between the events. The word “or” indicates union so the probability we seek is P(R ∪ B). Using the union formula above with the events R and B gives
From the problem statement, we see one probability uses the word “and”. This signals intersection so we know P(R ∪ B) = 0.1. The other probabilities in the problem give us P(R) = 0.4 and P(B) = 0.55. The union formula gives us
b. What is the probability that neither the rear brakes need adjusting nor the rear brakes need adjusting?
Solution The key to solving this problem is to associate the event “neither the rear brakes need adjusting nor the rear brakes need adjusting” with the events R and B. This event is the compliment of the event in part a, P((A ∪ B)’). This means we can use the compliment formula,
to find the probability that neither the rear brakes need adjusting nor the rear brakes need adjusting. Putting in the answer from part a gives
