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UI STAT 251 Spring 2011
Quiz 8
NAME:
UI #:
You play two games against the same opponent. The probability you win the first game is 0.6. If you lose the first game,
the probability that you also lose the second one is 0.7. However, if you win the first game, the probability you also win
the second is 0.8.
a. Find the probability of winning both games.
𝑃(𝑤𝑖𝑛 𝑏𝑜𝑡ℎ) = 𝑃(𝑤𝑖𝑛 1𝑠𝑡) ∗ 𝑃(𝑤𝑖𝑛 2𝑛𝑑) = 0.6 ∗ 0.8 = 0.48
b. Find the probability of losing both games.
𝑃(𝑙𝑜𝑠𝑒 𝑏𝑜𝑡ℎ) = 𝑃(𝑙𝑜𝑠𝑒 1𝑠𝑡) ∗ 𝑃(𝑙𝑜𝑠𝑒 2𝑛𝑑) = 0.4 ∗ 0.7 = 0.28
c. Find the probability of winning the second game given that you won the first game.
𝑃(𝑤𝑖𝑛 2𝑛𝑑|𝑤𝑖𝑛 1𝑠𝑡) =
𝑃(𝑤𝑖𝑛 𝑏𝑜𝑡ℎ) 0.48
=
= 0.8
𝑃(𝑤𝑖𝑛 1𝑠𝑡)
0.6
d. Find the probability of winning the second game.
𝑃(𝑤𝑖𝑛 2𝑛𝑑) = 𝑃(𝑤𝑖𝑛 1𝑠𝑡 𝑡ℎ𝑒𝑛 𝑤𝑖𝑛 2𝑛𝑑) + 𝑃(𝑙𝑜𝑠𝑒 1𝑠𝑡 𝑏𝑢𝑡 𝑤𝑖𝑛 2𝑛𝑑) = 0.6 ∗ 0.8 + 0.4 ∗ 0.3 = 0.6
e. Are the two games independent? Explain.
f.
𝑃(𝑤𝑖𝑛 2𝑛𝑑) ≠ 𝑃(𝑤𝑖𝑛 2𝑛𝑑|𝑤𝑖𝑛 1𝑠𝑡), so the two games are not independent.
Let random variable X be the number of games you win. What are the possible values of X?
The possible values for X are 0, 1 and 2.
g. Find the probability distribution model for X (i.e., find the probability for each possible value of X)
X
P
0
0.28
1
0.24
2
0.48
𝑃(𝑋 = 0) = 𝑃(𝑙𝑜𝑠𝑒 𝑏𝑜𝑡ℎ) = 0.28, 𝑃(𝑋 = 2) = 𝑃(𝑤𝑖𝑛 𝑏𝑜𝑡ℎ) = 0.48, 𝑃(𝑋 = 1) = 1 − 0.28 − 0.48 = 0.24
h. What is the expected value of X?
𝐸(𝑋) = ∑ 𝑋 ∗ 𝑃(𝑋) = 0 ∗ 0.28 + 1 ∗ 0.24 + 2 ∗ 0.48 = 1.2
i.
What is the variance of X?
𝑉𝑎𝑟(𝑋) = ∑(𝑋 − 𝐸(𝑋))2 ∗ 𝑃(𝑋) = (0 − 1.2)2 ∗ 0.28 + (1 − 1.2)2 ∗ 0.24 + (2 − 1.2)2 ∗ 0.48 = 0.72
j.
What is the standard deviation of X?
𝑆𝐷(𝑋) = √0.72 = 0.85