You play two games against the same opponent. The probability you win the first
ID: 3066206 • Letter: Y
Question
You play two games against the same opponent. The probability you win the first game is
0.60.6. If you win the first game, the probability you also win the second is 0.30.3. If you lose the first game, the probability that you win the second is
0.20.2. Complete parts a) through e).
a) Are the two games independent? Explain your answer.
A.No; the outcome of the first game determines the probability of winning the second game.
B.Yes; the outcome of the first game has no impact on the second game.
C.No; no events are independent.
D.Yes; all events are independent.
b) What's the probability you lose both games?----------
(Type an integer or a decimal.)
c) What's the probability you win both games? ----------
(Type an integer or a decimal.)
d) Let random variable X be the number of games you win. Find the probability model for X.
x
0
1
2
P(x)
----------
----------
---------
e) Find the expected value and standard deviation of X.
E(X)equals=--------
(Type an integer or a decimal.)
SD(X)equals= --------
(Round to three decimal places as needed.)
x
0
1
2
P(x)
----------
----------
---------
Explanation / Answer
a) Are the two games independent? Explain your answer.
Answer : Option A.
No; the outcome of the first game determines the probability of winning the second game.
b) What's the probability you lose both games
P(lose 1st)=1 -0.60=0.40
P(lose 2nd) = 1-0.20 =0.80
Hence, P(lose both)=P(lose 1st)*P(lose 2nd)
= 0.40*0.80
The probability you lose both games= 0.32
c) What's the probability you win both games?
P(win both)=P(win 1st)*P(win 2nd)
= 0.60*0.3
The probability you win both games =0.18
d) Let random variable X be the number of games you win. Find the probability model for X.
x
0
1
2
p(X)
0.32
0.50
0.18
P(X=0)=P(loseboth)=0.32
P(X=2)=P(win both)=0.18
P(X=1)=1-0.32-0.18=0.50
e) Find the expected value and standard deviation of X.
E(X)=X*P(X)=0*0.32+1*0.50+2*0.18=0.86
expected value =0.86
Var(X)=[(X-E(X))]^2 *P(X)
=[(0-0.86)^2*0.32]+[(1-0.86)^2*0.5]+[(2-0.86)^2*0.18]
=0.87028
standard deviation of X= sqrt (V(X))= 0.933
x
0
1
2
p(X)
0.32
0.50
0.18
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