91Ó°ÊÓ

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.

We need to find that the probability that the present is upstairs .

The solution is,

$A=$event that present was hidden by mom

$B=$event that present was hidden by dad.

$E=$event that she hides in upstairs.

$F=$event that she hides in downstairs.

Then,

$P\left(A\right)=.6$

$P\left(B\right)=.4$

$P(Eâˆ\pounds A)=.70$

$P(Fâˆ\pounds A)=.30$

$P(Eâˆ\pounds B)=.5$

$P(Fâˆ\pounds B)=.5$

The probability that the present is upstairs is,

$P\left(E\right)=P\left(A\right)P(Eâˆ\pounds A)+P\left(B\right)P(Eâˆ\pounds B)$

$=\left(.6\right)\left(.70\right)+\left(.4\right)\left(.5\right)$

$=.42+.20$

$=.62$

The probability that the present is upstairs is$0.62$.

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.

We need to find that is the probability it was hidden by dad

The probability that it was hidden by dad if it is downstairs will be,

$P(Bâˆ\pounds F)=\frac{P(Fâˆ\pounds B)P\left(B\right)}{P(Fâˆ\pounds A)P\left(A\right)+P(Fâˆ\pounds B)P\left(B\right)}$

$=\frac{\left(.5\right)\left(.4\right)}{\left(.30\right)\left(.6\right)+\left(.5\right)\left(.4\right)}$

$=\frac{.20}{.38}$

$=\frac{10}{19}$

$≈0.5263$

The probability it was hidden by dad if the given are down stairs is$0.5263$.