A penny and a nickel are tossed. Both are fair coins. Let X = 1 if the penny comes up heads, and let X = 0 otherwise. Let Y = 1 if the nickel comes up heads, and let Y = 0 otherwise. Let Z = 1 if both the penny and nickel come up heads, and let Z = 0 otherwise.

a. Let px denote the success probability for X. Find px.

b. Let pY denote the success probability for Y. Find pY.

c. Let pz denote the success probability for Z. Find pz.

d. Are X and Y independent?

e. Does pz=pxpy?

f. Does Z = XY? Explain.

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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