9.1. Let Xi for i = 2, . . . 7 be the random variable P
equal to 1 if there is a
7
stop on the floor number i and to 0 otherwise. Then i=2 Xi is the number
P7
of stops. Consequently, the expected number of stops is equal to i=2 EXi .
Let us find distribution of Xi . The probability P (Xi = 0) is equal to (5/6)20 ,
since each person independently with probability 5/6 does not stop on the ith
P7
floor. Consequently, EXi = P (Xi = 1) = 1 − (5/6)20 , hence
i=2 EXi =
6(1 − (5/6)20 ) ≈ 5.84.
9.2. Let X1 , X2 , X3 , X4 be the random variables, equal to 1 if the corre(39
5)
sponding suit appears, and 0 otherwise. Then P (Xi = 0) = 52
= 39!47!
34!52! =≈
(5)
0.22. It follows that EXi = P (Xi = 1) ≈ 0.78. Consequently, the expected
number of different suits is approximately 4 · 0.78 = 3.11.
9.3. E(3X + 4Y − 5) = 3EX + 4EY − 5 = 3 + 8 − 5 = 6.
V ar(3X + 4Y − 5) = 9V ar(X) + 16V ar(Y ) = 27 + 16 = 43.
9.4. cov(X, Y ) = E(XY ) − EX · EY . Expectation of X is 0. The product
XY is equal to X 3 and its expectation is 0. It follows that cov(X, Y ) = 0.
X and Y are not independent, since one is a function of the other, and they
are not constant. (In particular, P (X = 1, Y = 0) is zero, while P (X = 1) and
P (Y = 0) are both different from zero.)
9.5. We have EX = 2 and V ar(X) = 4 · 1/2 · 1/2 = 1. By Chebyshev’s
inequality,
V ar(x)
= 1/4.
P (|X − EX| ≥ 2) ≤
22
Exact probability is equal to P (X = 4) + P (X = 0) =
1
1
16
+
1
16
= 18 .