Question #16852

Nancy is throwing a standard six sided die. If she throws a 4 on her first throw, what is the probability that her next throw will NOT be. A 4?

Expert's answer

Let X be the result of first throwing, and Y be the result of thesecodn

throwing.

We will assume that X and Y are independent, since we have no

information.

This means that

p(X=k,Y=m) = p(X=k) *

p(Y=m)

Since die is standard six sided,

p(X=k) = p(Y=m) =

1/6.

Hence

p(X=k,Y=m) = 1/6 * 1/6 = 1/36

for any k,m.

We

should compute

p(X=4, Y<>4) = p(X=4, Y=1) + p(X=4, Y=2) + p(X=4, Y=3)

+ p(X=4, Y=5) + p(X=4, Y=6) =

= 5 * 1/36 =

5/36.

Answer: 5/36

throwing.

We will assume that X and Y are independent, since we have no

information.

This means that

p(X=k,Y=m) = p(X=k) *

p(Y=m)

Since die is standard six sided,

p(X=k) = p(Y=m) =

1/6.

Hence

p(X=k,Y=m) = 1/6 * 1/6 = 1/36

for any k,m.

We

should compute

p(X=4, Y<>4) = p(X=4, Y=1) + p(X=4, Y=2) + p(X=4, Y=3)

+ p(X=4, Y=5) + p(X=4, Y=6) =

= 5 * 1/36 =

5/36.

Answer: 5/36

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