Answer:
a) px = P(X=1) = 1/2
b) py = P(Y=1) = 1/2
c) pz = P(Z=1) = 1/4
d) X and Y are independent events as they do not depend on each other to occur.
e) yes, pz = px py
f) yes, Z = XY
Step-by-step explanation:
The sample space for tossing both a penny and a nickel includes HH, HT, TH, TT
n(sample space) = 4
a) Let px denote the success probability for X. Find px.
X = 1 if the penny comes up heads, and X = 0
px = success probability for X and that is p(X=1)
p(X=1) = 2/4 = 1/2 (out of the four possible outcomes, only 2 have the penny come up heads; HH and HT)
b) py = P(Y=1) = 2/4 = 1/2 (out of the four possible outcomes, only 2 have the nickel come up heads; TH and HH)
c) pz = P(Z=1) = 1/4 (out of the four possible outcomes, only one has the penny and nickel come up heads; HH)
d) X and Y are independent events as they do not depend on each other to occur. Occurrence of a penny turning up heads, doesn't affect the probability of a nickel turning up heads.
Mathematically, for two independent events,
P(X n Y) = P(X) × P(Y) = (1/2) × (1/2) = 1/4 = P(Z)
e) pz = P(Z) = the probability of both penny and nickel turn up heads
And since we've established that X, probability of a penny head is independent of getting a nickel head, Y.
pz = px py = (1/2)(1/2) = 1/4 (proved)
f) To prove Z = XY
when X = 1, And Y = 1, that is, HH
XY = 1×1 = 1 and Z = 1 too since HH is its conditiin to be a 1. Hence Z = XY = 1 here.
when X = 1 and Y = 0, that is, HT
XY = 1×0 = 0 and Z = 0, since any deviation from both heads (HH) is a 0 for Z. Hence, Z = XY = 0 here.
when X = 0 and Y = 1, that is, TH
XY = 0×1 = 0 and Z = 0, since any deviation from both heads (HH) is a 0 for Z. Hence, Z = XY = 0 here too.
when X = 0 and Y = 0, that is, TT
XY = 0×0 = 0 and Z = 0, since any deviation from both heads (HH) is a 0 for Z. Hence, Z = XY = 0 here too.
Since Z = XY for all the cases, Z is indeed equal to XY.
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