Stat 331/Elec 331, Solutions to Homework 2
1a. P(A) = 1/2, P (B) = 2/3, P (C) = 3/36 and P(A∩B∩C) = 1/36.
b. No, since for instance Aand Care not independent; P(A∩C) = 3/36 6=
P(A)P(C) (actually, if Coccurs, Amust also occur). That P(A∩B∩C) =
P(A)P(B)P(C) is not enough for independence.
2a. Let Ajbe the event that we get the ace of spades in the jth draw. Since
we draw with replacement and independently, the event that we get the ace
of spades for the first time in the kth draw is the event Ac
1∩Ac
2∩...∩Ac
k−1∩Ak
and by independence the probability of this is P(Ac
1)·... ·P(Ac
k−1)P(Ak) =
(51/52)k−11/52 (a geometric distribution with p= 1/52).
b. The probability that the ace of spades appears in the first trial is 1/52,
hence P(X= 1) = 1/52. That it appears in the second trial means that we
must draw something else in the first trial, which has probability 51/52, and
then draw the ace of spades from the remaining 51 cards which has prob-
ability 1/51. Hence P(X= 2) = 51/52 ·1/51 = 1/52 (formally we have
used the formula P(C∩D) = P(D|C)P(C) here). Similarly we will have
P(X=k) = 1/52 for k= 1,2, ..., 52.
Don’t confuse conditional and unconditional probabilities here: the probabil-
ity that the ace of spades appears in the 51st trial is 1/52 but the conditional
probability that it appears in the 51st trial given that it has not appeared
before is 1/2.
3. For each function f, check that f≥0 and R∞
−∞ f(x)dx = 1.
a. This is a pdf. To compute the distribution function, first take −1≤x≤
0. Then Rx
−1f(t)dt =Rx
−1−tdt =−[t2/2]x
−1= 1/2(1 −x2). For 0 ≤x≤1 we
have and Rx
−1f(t)dt = 1/2 + Rx
0tdt = 1/2(1 + x2). Hence
F(x) = (1/2(1 −x2) if −1≤x≤0
1/2(1 + x2) if 0 ≤x≤1
b. This is not a pdf since it is negative between 0 and 1 (that it has integral
equal to 1 does not help in this case).
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