IE 111
Solutions
9:00 Section
Exam 2
Fall 2009
Question 1
Three
flips of a biased coin are conducted in which the probability of a head is 0.4.
Let X
1
= the number of heads in the
first
flip (i.e. flip number 1).
Let X
2
= the number of heads in the
last
two flips (i.e. flip numbers 2 and 3).
Let X
3
= the number of heads in the
last
flip (i.e. flip number 3).
a)
Find P(X
1
+X
2
<2)
This happens exactly when there is 0 or 1 head on the three flips together. The probability
is thus
0.6
3
3
×
0.6
2
×
0.4
=
0.648
.
b)
Find E(X
1
+X
2
+X
3
)
By the properties of the mean, we can write E(X
1
+X
2
+X
3
)=EX
1
+EX
2
+EX
3
, which we can
compute by terms. EX
1
=0.4=EX
3
, and EX
2
=0.8, since X
2
is a binomial random variable
with N=2, p=0.4, and the expected value is Np. So the some of the three terms is 1.6.
c)
Find P(X
1
+X
2
+X
3
<3)
There are only 8 possibilities for the coin flips, so we can lsit all of them together with
the value of X
1
+X
2
+X
3
and the probability of that combination.
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 Spring '07
 Storer
 Probability, Probability theory, Cumulative distribution function, CDF, 30 seconds, 55%

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