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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,

$P\left(E^câˆ\pounds A\right)=.3$

$P\left(E^câˆ\pounds A^c\right)=.1$

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

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,

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

$=\left(1−P\left(E^câˆ\pounds A\right)\right)P\left(A\right)+\left(1−P\left(E^câˆ\pounds A^c\right)\right)P\left(A^c\right)$

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

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

$=.76$

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,

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

$=\frac{\left(1−P\left(E^câˆ\pounds A\right)\right)P\left(A\right)}{P\left(E\right)}$

$=\frac{(1−.3)\left(.7\right)}{.76}$

Therefore,

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

$=.644736842$

$=\frac{49}{76}$

$≈0.64474$

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