The question is incomplete. Here is the complete 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 probability it was hidden by dad?
Answer:
(a) 0.62
(b) 0.526 (correct to 3 decimal places)
Step-by-step explanation:
Let Probability of present hidden by mom represented by P(M)
Let Probability of present hidden by father represented by P(F)
Let Probability of present hidden upstairs by mother represented by P(Um)
Let Probability of present hidden downstairs by mother represented by P(Dm)
Let Probability of present hidden upstairs by father represented by P(Uf)
Let Probability of present hidden downstairs by father represented by P(Df)
P(M) = 0.6
P(F) = 0.4
P(Um) = 0.7
P(Dm) = 0.3
P(Uf) = 0.5
P(Df) = 0.5
Part (a)
Present is upstairs can have two possibilities
Hidden by mother and hidden upstairs = P(M) x P(Um)
= 0.6 x 0.7
= 0.42
Or
Hidden by father and hidden upstairs = P(F) x P(Uf)
= 0.4 x 0.5
= 0.2
Final answer = 0.42 + 0.2
= 0.62
Part (b)
It is a question of conditional probability
Probability of present hidden downstairs = 1 – Probability of present hidden upstairs (found in part (a))
= 1 – 0.62
= 0.38
Probability of present hidden downstairs by father= P(F) x P(Df)
= 0.4 x 0.5
= 0.2
Conditional Probability equation
<em>P(F|Downstairs) = Probability of present hidden downstairs by fathe</em>r<em> / Probability of present hidden downstairs
</em>
P(F|Downstairs) = 0.2/0.38
P(F|Downstairs) = 0.526 (correct to 3 decimal places)
