The following shows the results of a wine tasting survey of 100 people: 96 like wine A, 93 like wine B, 96 like wine C, 92 like A and B, 91 like B and C, 93 like A and C, 90 like all three wines - Edexcel - A-Level Maths Statistics - Question 5 - 2008 - Paper 1

The number of people who like none of the wines is 1. To find the probability, we have:

$P(None) = \frac{n(None)}{100} = \frac{1}{100} = 0.01$

The number of people who like wine A but not wine B is 4. Therefore, the probability is given by:

$P(A \cap \neg B) = \frac{4}{100} = 0.04$

The total number of people liking wine C is 96. Those liking any wine excluding C are:

Total = 100 - n(C) = 100 - 96 = 4.

Thus, the probability is:

$P(\neg C) = \frac{4}{100} = 0.04$

To find those who like exactly two kinds:

• A and B only: 2
• A and C only: 3
• B and C only: 1

Total = 2 + 3 + 1 = 6. Thus, the probability is:

$P(Exactly\; 2) = \frac{6}{100} = 0.06$

Given that a person likes wine A, we need to find the intersection with C. We know 90 people like all wines, plus the others liking A-C. Thus, we have associated values. Therefore, the probability is:

$P(C | A) = \frac{P(A\cap C)}{P(A)} = \frac{93}{96} \approx 0.96875$. This means, a person who likes A, likely also likes C.