Answer:
The probability that is raining given that Randy is late to class is P(R|L)=0.175.
Step-by-step explanation:
We can apply the Bayes theorem to solve this question.
Being
L: Randy is late to class
R: It is raining
nL: Randy is on time to class
nR: It is not raining
We know:
P(L | R)= 1-0.7=0.3
P(L | nR) = 1-0.9=0.1
P(R) = 0.2 (probability of Randy going to class given that is raining)
P(nR) = 0.8
We have to calculate P(R | L): probability that is raining given that Randy is late to class.
If we apply the Bayes theorem, we have:
[tex]P(R | L)=\dfrac{P(L|R)P(R)+P(L|nR)P(nR)}{P(L|R)+P(L|nR)}=\dfrac{0.3\cdot0.2+0.1\cdot0.1}{0.3+0.1}\\\\\\P(R | L)=\dfrac{0.06+0.01}{0.4}=\dfrac{0.07}{0.4}=0.175[/tex]
The probability that is raining given that Randy is late to class is P(R|L)=0.175.
