Given ind. prob's, find prob. that he does and she doesn't

[Navyguy]

The probability that Andy studies math over break is 0.3 and the probability that Betty independently studies over break is 0.4. Find the probability Andy studies over break and Betty doesn't.

The percentage is messing me up when I try to set up the problem. I have tried multiplying and dividing, and only came up with 0.12. Please show me a formula that I can use to set this problem up. The list of possible answers is as follows:

1. 0.12
2. 0.42
3. 0.18
4. 0.46

Thanks for any help given.

 

[pka]

Suppose that we have two independent events each of which has non-zero probability. Then \(\displaystyle P\left( {A|B} \right) = P(B|A) = P(A)P(B)\)
It is a theorem that if A & B are independent then so are each of these pairs:
\(\displaystyle \left\{ {A,\overline B } \right\},\;\left\{ {B,\overline A } \right\},\;\left\{ {\overline A ,\overline B } \right\}\quad \overline B \mbox{ is the complement of B}\)
\(\displaystyle \begin{array}{l}
P(B) = 0.4\quad \Rightarrow \quad P\left( {\overline B } \right) = 1 - 0.4 = 0.6 \\
P\left( {A \cap \overline B } \right) = P\left( A \right)P\left( {\overline B } \right) \\
\end{array}\)

 

[Navyguy]

okk let me see if I got this much of it

P(A)=0.3 P(A)= 1-0.3=0.7, is this part of the formula you are showing me. Is the next step adding the two or is that wrong.

 

[pka]

You are simply looking for;
\(\displaystyle P(A \cap \overline B ) = P(A)P(\overline B ).\)
Andy studies and Betty does not.