Back to: Chapter 6: Probability

## 10 thoughts on “Q2 | Solving Word Problems (Preview)”

1. Is there a way to know which value is A and which value is B in an equation? Also, in the probability tree A(adopted) was A while R (received) was B and in the multiplication rule and joint prob, they reversed. Do you mind clarifying why that happened? Thanks

1. Good question – one that I’m sure a lot of students share.
Each time you use a probability formula, you must decide yourself which outcome to assign to A and which to assign to B. The wording of the question almost always tells you how to properly assign them. The ONLY restriction is that the meaning of A and B must remain consistent through the formula. So, if you are using a formula like:

P(A or B) = P(A) + P(B) – P(A|B)

and you have the outcomes:

Then there are a number of ways that you could assign A and B – none of them are the “only” correct choice. However, if the question asks… “Find the probability of selecting either Adopted OR received”.. then it makes sense to assign A and B as follows:

P(A or B) = P(A) + P(B) – P(A|B)

…and keeping the assignment consistent on the other side of the equation gives us…

Summary:
=========
– The wording of the question tells you how to assign A and B
– You must be consistent throughout the entire formula

2. Hi Jason,

if A and B are independent events with P(A) = 0.60 and P(B)= 0.70, then the probability that A occurs or B occurs or both occur is:

1. “A occurs or B occurs or both” means that we are looking for P(A or B).

The formula is:
P(A or B) = P(A) + P(B) – P(A and B)

We have:
P(A) = 0.60
P(B) = 0.70

We just need P(A and B).

The formula is:
P(A and B) = P(A | B) x P(B)

…but independent events means that:

P(A | B) = P(A)

Therefore, P(A and B) = P(A | B) x P(B) = P(A) x P(B) = (0.60)(0.70) = 0.42

So P(A or B) is…

P(A or B) = P(A) + P(B) – P(A or B) = (0.60) + (0.70) – (0.42) = 0.88

3. how can you know if to solve a question from contingency table or tree diagram(conditional)?

1. Tree diagrams aren’t just for conditional probabilities. If you flip a coin twice in a row, none of the outcomes have conditional probabilities, but the tree is still good at helping you sort out all of the joint probabilities. That’s really what tree diagrams are for – They allow you to easily organize and calculate all of the joint probabilities. This is handy sincex there can be a lot of joint probabilities (there is one for each outcome).

Conditional tables require that you already know the joint probabilities in order to set up the table. The main purpose of the conditional table is to organize all of these joint probabilities, and also to assist in the calculation of the marginal probabilities.

So the best strategy depends on what you have been given in the question:

Given: Joint probabilities >> Go right to the conditional table

Given: No joint probabilities >> Draw the tree >> determine the joint probabilities >> Create the contingency table

4. How do you know to put P(R|A) = P(R and A) / P(A) when the question asks “What is the probability that a professor who adopts the book has received the advertising material?” – (Question A)

Since they ask about the professor adopting the material first and then ask if they received the material wouldn’t it be P(A|R) = P(A and R) / P(R).

“A” obviously being that the professor adopted the material and “R” being that they received it.

1. Conditional probabilities can be thought of as

P(what happened | to who)

The “who” in this case is the professor who adopts the book.

– Hope this helps!

5. I’m very confused. at 27:27 the formula is p(a and b)= p(r)p(a|r). i don’t under stand how P(a|r) is equal to 0.28. isn’t 0.28 the p(R and A).

1. Hello Venisha,

I believe you mean:

P(A and R) = P(R) P(A | R)

From the setup:

“twenty eight percent of the professors who received this material adopted the book”

If you find that a probability only applies to a subgroup (not everyone), then it is a conditional probability. The given part of the conditional will be who the probability applies to.