Question

Suppose we have the following information about how many Canadians were watching these events:

56% watched Ski Jumping (SK)
42% watched Snowboarding (SB)
67.44% watched Ski Jumping or Speed Skating (SS)
55.65% watched Snowboarding or Speed Skating
35.86% watched Ski Jumping and Snowboarding
14.56% watched Ski Jumping and Speed Skating
8.29% watched all three events.
We will randomly select one Canadian.

What is the probability that the selected person didn't watch any of the three events?

Answer 1

Let:
\(K\)=ski jumping
\(B\)=snow boarding
\(S\)=speed skating

Then, we want to know the prob. that a person didn't watch any of the 3 events. This can be translated as "Didn't watch \(K\), and didn't watch \(B\), and didn't watch \(S\)". In probability terms, you want \(P(\overline{K} \cap \overline{B} \cap \overline{S})\) which equals to: \(P(\overline{K \cup B \cup S})\) by DeMorgan's Law.

What we are given:
\(P(K)=0.56\)
\(P(B)=0.42\)
\(P(K \cup S)=0.6744\)
\(P(B \cup S)=0.5565\)
\(P(K \cap B)=0.3586\)
\(P(K \cap S)=0.1456\)
\(P(K \cap B \cap S)=0.0829\)

By the inclusion/exclusion principle:
\[P(K \cup B \cup S)=P(K)+P(B)+\color{red}{P(S)}-P(K \cap B)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~-P(K \cap S)-\color{red}{P(B \cap S)}+P(K \cap B \cap S)\]

In red is the information not directly given, which you can find with other probability rules.

Answer 2

\(P(B \cap S)\) can be found from: \(P(B \cup S)=P(B)+P(S)-P(B \cap S)\)
Although you're still missing P(S), which could be found from:

\(P(K \cup S)=P(K)+P(S)-P(K \cap S)\)