# multiplication rule with random events

Exercise from mathint.com:

"The probability that a student passes Mathematics is 2/3 and the probability that he passes English is 4/9 . If the probability that he will pass at least one subject is 4/5, what is the probability that he will pass both subjects? (We assume it is based on probability only.)"

My guess would be to apply the multiplication rule because these are two independent events. So the solution would be 2/3 * 4/9 = 8/27. But, according to the solution provided on the website, my solution is wrong. And while I understand the explanation they provide, I do not understand why my version is incorrect.

Is it because the probability of passing a Math and English exam is unlike the probability of drawing two cards from a deck with replacement? Eg. a deck will always have the same 52 cards, but a Math and EN exam can differ in its content, difficulty, length of time required, etc, so multiplying the probabilities of passing them makes no sense, and is like comparing apples and oranges?

Thank you for your help.

• What tells you these events are independent?
– whuber
Oct 4, 2017 at 20:43
• Because I assume that one's performance in Math does not influence ones performance in English, or vice versa. Wouldn't you agree? Oct 4, 2017 at 20:52
• It doesn't matter whether I agree or not: independence is not a fact that's given in the question. Indeed, if the outcomes were independent, then the chance of failing both subjects would be $(1-2/3)\times(1-4/9)=5/27$. Is that consistent with a $4/5$ chance of passing at least one subject??
– whuber
Oct 4, 2017 at 21:06
• OK, I over-complicated this. So basically the answer is that without knowing the addition rule, there is no way to solve this exercise, right? Oct 4, 2017 at 21:34
• "Yes, if I multiply the two sets, then...." You have a strange hang-up where you have the urge to multiply things. Resist using the word (or performing the action) unless absolutely necessary Oct 6, 2017 at 4:02

## 2 Answers

The addition rule, solving for $P(A \cap B)$:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) \equiv P(A \cap B) = P(A) + P(B) - P(A \cup B)$$

# Implementation in R
A       = 2/3  # P that he passes math
B       = 4/9  # P that he passes English
A_or_B  = 4/5  # P that he passes at least one subject
A_and_B = NA   # P that he passes both subjects

A_and_B = A + B - A_or_B
A_and_B
> 0.3111111


Whoever asked this question might be hoping you use the rule of addition which states: p(A or B) = p(A) + p(B) - p(A and B)

Events need not be mutually exclusive. You have p(A), p(B), and p(A or B) so you can solve for p(A and B).