Q.3.33 - A First Course in Probability

Question

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?

Step-by-Step Solution

Answer

(a) The probability that Joe is early tomorrow is $.76$.

(b) The conditional probability that it rained is $0.64474$.

Step 1: Given information (Part a)

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. 

Step 2: Solution (Part a)

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.

$=1-.7$

$=.3$

Step 3: final solution (Part a)

The probability that Joe is early tomorrow will be,

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

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

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

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

$=.76$

Step 4: Final answer (Part a)

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

Step 5: given information (Part b)

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.

Step 6: Solution (Part b)

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,

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

$=.644736842$

$=\frac{49}{76}$

$\approx 0.64474$

Step 7: Final answer (part b)

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