probability

Question

With probability .6, the present was hidden by
mom; with probability .4, it was hidden by dad.
When mom hides the present, she hides it upstairs
70 percent of the time and downstairs 30 percent
of the time. Dad is equally likely to hide it upstairs
or downstairs.
(a) What is the probability that the present is
upstairs?
(b) Given that it is downstairs, what is the proba-bility it was hidden by dad?

Answer 1

p(mom,upstairs)=0.6×0.7=0.42;

p(mom,downstairs)=0.6×0.3=0.18;

p(dad,upstairs)=0.4×0.50=0.20;

p(dad,downstairs)=0.4×0.50=0.20.

[Total probability=0.42+0.18+0.20+0.20=1 (all outcomes covered).]

(a) p(upstairs)=0.42+0.20=0.62;

(b) p(downstairs)=0.18+0.20=0.38 (both parents);

p(dad|downstairs)=0.20/0.38=0.53 approx.

[Total probability=0.62+0.38=1 (all outcomes covered).]

EXAMPLE

There are 100 presents.

Mom hides 60, dad hides 40.

Mom hides 42 (70% of 60) upstairs and 18 downstairs (30% of 60).

Dad hides 20 (50% of 40) upstairs and 20 downstairs (50% of 40).

There are 38 presents downstairs and 62 upstairs.

Out of the 38 downstairs, dad has hidden 20. This is a probability of 20/38=10/19.