Question

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?

Answer 1

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