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The problem is:

A certain shop repairs both audio and video components. Let A denote the event that the next component brought in for repair is an audio component, and let Bbe the event that the next component is a compact disc player (so the event B is contained in A). Suppose that $P(A)=.6$ and $P(B)=.05$ What is ? $P(B|A)$

If I knew $P(A \cap B)$, then I could calculate $P(A \cup B)$. I thought by introducing a new set, one that is the complement of $A$, I could someone how make this problem work. I introduced the set $V$, which is the event someone brings in an video component, and then $P(A) + P(V) = 1 \implies P(V) = .4$. However, this didn't seem--ostensibly at least--to provide any help. Could someone prod me along?

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up vote 1 down vote accepted

Since $B$ is contained in $A$, $A\cap B=B$. So you do know $P(A\cap B)$.

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So, if I were to draw a Venn Diagram of $A$ and $B$, the circle representing $B$ would be completely inside of the circle representing $A$? – Mack Feb 5 '13 at 0:28
@Eli: All CD players count as audio equipment, don't they? – Henning Makholm Feb 5 '13 at 0:30
I suppose they do. Thank you for aiding me in this simple err on my part – Mack Feb 5 '13 at 0:33

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