1. ## Quick probability question

Hello everyone,

I'm finishing up some of my stat HW right now and I have hit a wall. There are two questions that are on the tip of my tongue and I can't answer them! It's maddening!

Anyway, here is the the first question...in question:

Suppose the P(E) = .6 , P(F) = .3, and that E and F are independent events. What is the probability that the student knows the answer to the first question and does NOT know the answer to the second question?

(E=prob of first question and F= prob of second)

I really feel like the solutions are obvious, but nothing is coming to me.

Question two is:

P(E) = .6 and P(F) = .5 and P(E and F) = .4

Given the above, what is the probability that the student knows the answer to EXACTLY one of the questions?

Thanks!

2. Hello Nohg

Welcome to Math Help Forum!
Originally Posted by Nohg
Hello everyone,

I'm finishing up some of my stat HW right now and I have hit a wall. There are two questions that are on the tip of my tongue and I can't answer them! It's maddening!

Anyway, here is the the first question...in question:

Suppose the P(E) = .6 , P(F) = .3, and that E and F are independent events. What is the probability that the student knows the answer to the first question and does NOT know the answer to the second question?

(E=prob of first question and F= prob of second)

I really feel like the solutions are obvious, but nothing is coming to me.

Question two is:

P(E) = .6 and P(F) = .5 and P(E and F) = .4

Given the above, what is the probability that the student knows the answer to EXACTLY one of the questions?

Thanks!
1) If $\displaystyle p(F)=0.3$, then $\displaystyle p(F')$ (= the probability that the student does not know the answer to the second question) $\displaystyle = 1-0.3=0.7$.

And if the events are independent, $\displaystyle p(E \cap F')=p(E)p(F') = 0.6 \times 0.7 = 0.42$.

2) $\displaystyle p(E \cup F) = p(E) + p(F) -p(E\cap F) = 0.6+0.5-0.4=0.7$

But this is an inclusive OR; in other words, it is the probability that the student knows the answer to question 1 or question 2 or both. So if we want the probability that he knows the answer to exactly one question, we subtract from this the probability that he knows the answer to both. Thus:

$\displaystyle p(E$ or $\displaystyle F$ but not both$\displaystyle )=p((E \cup F) \cap (E\cap F)') = 0.7-0.4=0.3$