Quick probability question

Nohg · #1 $Old$ November 5th, 2009, 01:59 PM

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!

Grandad · #2 $Old$ November 6th, 2009, 09:12 AM

Hello Nohg

Welcome to Math Help Forum!

Quote:

Originally Posted by Nohg $View Post$

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

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

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

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:

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

Grandad