# Find the probability of these events using basic probability principles

Suppose there are two types of Penn State Sports fans: Rabid (R), and Casual (R'). These fans either view a big game Live (L) or at Home (L'). Suppose 15% of Penn State fans are both Casual and watch the games at Home. If a fan watches at Home, the probability that he/she is Casual is 80%. Finally, suppose 10% of fans are Rabid.

a. What is the probability of any fan watching the game at Home?

b. If a fan watches the game Live, what is the probability that they are Rabid?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Suppose there are two types of Penn State Sports fans: Rabid (R), and Casual (R'). These fans either view a big game Live (L) or at Home (L'). Suppose 15% of Penn State fans are both Casual and watch the games at Home. If a fan watches at Home, the probability that he/she is Casual is 80%. Finally, suppose 10% of fans are Rabid.

R = Rabid P(R) = .10

R'= Casual P(R') = 1 - .10 = .90

L = Live

L' = Home

P(R'|L') = .80 P(R|L') = 1 - .80 = .20

a. What is the probability of any fan watching the game at Home?

P(R'|L') = .80

P(R'L') = .15 = P(L')P(R'|L') = P(L')(.80)

P(L') = (.15)/.80) = .1875

b. If a fan watches the game Live, what is the probability that they are Rabid?

Want P(R|L)

P(R) = P(RL) + P(RL') = P(RL) + P(L')P(R|L')

P(RL) = P(R) - P(L')P(R|L') = (.10) - (.1875)(.20) = .0625

P(L) = 1 - P(L') = 1 - .1875 = .8125

P(R|L) = P(RL)/P(L) = (.0625)/(.8125) = .0508

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