# 2.1. Let p be probability of winning. Then expectation of... p) + 3p = 4p − 1. It is equal... ```2.1. Let p be probability of winning. Then expectation of a win is (−1)(1 −
p) + 3p = 4p − 1. It is equal to zero when p = 1/4.
2.2. Probability of getting an odd man 2 &middot; 5 &middot; (1/2)5 . Here 2 stands for the
two possibilities of the result of toss for the odd man (heads or tails), 5 is the
number of cases who is the odd man. After we know this information, then we
know the result of toss for each of the players. Probability of getting a particular
result is (1/2)5 .
5
. The number of tosses till a
Thus, probability of getting an odd man is 16
successful finding an odd man has geometric distribution. The expectation of
the geometric distribution is inverse to the probability of success. Consequently,
the expected value of the number of tosses is 16
5 = 3.2.
2.3. Their goal is to get a baby whose sex is different from the sex of the
first child. Probability of this happening is 1/2 each time. So, after the first
child, the number of children they will have is geometrically distributed with
expected value 2. Consequently, the expected value of the number of children
they will have is 3.
2.4. Distribution of the number of days is P (X = 28) = 1/12, P (X = 30) =
= 365
4/12 = 1/3, and P (X = 31) = 7/12. The mean is EX = 28+4&middot;30+7&middot;31
12
12 .
EX 2 =
784 + 900 &middot; 4 + 961 &middot; 7
11, 111
=
,
12
12
hence
V ar(X) =
11, 111 3652
−
.
12
144
2.5. Variance is always non-negative, since it is equal to E(X − EX)2 . But
it is also equal to EX 2 − (EX)2 , and in this case it is 8 − 9 = −1, which is
impossible. 2.6. 19
7 .
1
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