Calculating probability from Venn diagram
I have this statistics problem where it's a Venn diagram relation. There's
88 people in total
21 belong to category A
17 belong to category B
11 belong to both A and B
39 belong to no category
I need to calculate the probability that when 3 people are selected at random from the whole sample, none belongs to category A.
When I do this, do I do
$88-(21+11)=56$
$(56/88)(55/87)(54/86)=(315/1247)$?
Do I need to subtract 1 from the denominator each time like I did to ensure the same person doesn't get chosen twice?
Is there another way to do this? This method seems odd to me because what if the question asks "30 people are chosen at random, what's the probability that no one is in group A"? Wouldn't that, then, take me ages to multiply all the probabilities together?
$\endgroup$ 21 Answer
$\begingroup$Are your sure it's 88 people altogether, not 66?
I get four disjoint categories $A \cap B = AB$ with 11, $BA^c$ with 6, $AB^c$ with 10, and $(A\cup B)^c$ with 39.
Then 11 + 6 + 10 + 39 = 66.
$\endgroup$ 2
