On rainy days, Joe is late to work with probability $.3$; on nonrainy days, he is late with probability $.1$. With probability $.7$, it will rain tomorrow.

(a) Find the probability that Joe is early tomorrow.

(b) Given that Joe was early, what is the conditional probability that it rained?

On rainy days, Joe is late to work with probability $.3$; on nonrainy days, he is late with probability $.1$. With probability $.7$

We need to find the probability that Joe is early tomorrow.

The solution is,

$A=$event that the rainy day.

$A^c=$event that the nonrainy day

$E=$event that Joe is early to work

$E^c=$event that Joe is late to work

Then,

$P\left(E^c\mid A\right)=.3$

$P\left(E^c\mid A^c\right)=.1$

$P\left(A\right)=.7$

So, $P\left(A^c\right)=1-P\left(A\right)$using the complementary rule.

$=1-.7$

$=.3$

The probability that Joe is early tomorrow will be,

$P\left(E\right)=P(E\mid A)P\left(A\right)+P\left(E\mid A^c\right)P\left(A^c\right)$

$=\left(1-P\left(E^c\mid A\right)\right)P\left(A\right)+\left(1-P\left(E^c\mid A^c\right)\right)P\left(A^c\right)$

$=(1-.3)\left(.7\right)+(1-.1)\left(.3\right)$

$=\left(.7\right)\left(.7\right)+\left(.9\right)\left(.3\right)$

$=.76$

The probability that Joe is early tomorrow is$.76$.

On rainy days, Joe is late to work with probability $.3$and on non-rainy days, he is late with probability $.1$. With probability $.7$.

We need to find that s the conditional probability that it rained.

The conditional probability that it rained if Joe is early will be,

$P(A\mid E)=\frac{P(E\mid A)P\left(A\right)}{P(E\mid A)P\left(A\right)+P\left(E\mid A^c\right)P\left(A^c\right)}$

$=\frac{\left(1-P\left(E^c\mid A\right)\right)P\left(A\right)}{P\left(E\right)}$

$=\frac{(1-.3)\left(.7\right)}{.76}$

Therefore,

$=\frac{\left(.7\right)\left(.7\right)}{.76}$

$=.644736842$

$=\frac{49}{76}$

$\approx 0.64474$

The conditional probability that it rained will be$0.64474$.