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 total number of people surveyed is 100. The number of people who like at least one wine can be found from the Venn diagram.

Calculating the total who like at least one wine:

$96 + 93 + 96 - 92 - 91 - 93 + 90 = 99$

Thus, those who like none of the wines:

$100 - 99 = 1$

The probability is:

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

To find the number of people who like wine A but not wine B:

This is simply the number of people who like only A plus those who like A and C (but not B):

$P(A \cap \neg B) = 1 + 3 = 4$

The probability is thus:

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

We need to find those who like wine A or wine B (but not wine C). From our counts:

• Only A: 1
• Only B: 0
• A and B (not C): 2

Thus:

$P(A \cup B \cap \neg C) = 1 + 0 + 2 = 3$

The probability is:

$P(any wine \neg C) = \frac{3}{100} = 0.03$

For exactly two kinds:

• A and B (not C): 2
• A and C (not B): 3
• B and C (not A): 1

Thus:

$P(exactly two) = 2 + 3 + 1 = 6$

The probability will be:

$P(exactly two) = \frac{6}{100} = 0.06$

Given that a person from the survey likes wine A:

• The number who like B and C is 1.
• Total who like A is 90.

The probability can be found using:

$P(C|A) = \frac{P(C \cap A)}{P(A)}\ = \frac{90}{90} = 1$

Thus, the probability is:

$P(C|A) = \frac{90}{100} = 0.90$