type the correct answer in each box. round your answers to the nearest hundredth. of a total of 100 buses…

type the correct answer in each box. round your answers to the nearest hundredth. of a total of 100 buses operated within a particular town, 48 usually run on time and 36 of these buses are owned by jerry. of the 52 buses that usually run behind schedule, only 7 are owned by jerry. the probability that a bus is owned by jerry, given that it runs on time, is %. the probability that a bus is owned by jerry, given that it runs late, is %.

type the correct answer in each box. round your answers to the nearest hundredth. of a total of 100 buses operated within a particular town, 48 usually run on time and 36 of these buses are owned by jerry. of the 52 buses that usually run behind schedule, only 7 are owned by jerry. the probability that a bus is owned by jerry, given that it runs on time, is %. the probability that a bus is owned by jerry, given that it runs late, is %.

Answer

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In terms of counts, if $n(A\cap B)$ is the number of elements in the intersection of $A$ and $B$ and $n(B)$ is the number of elements in $B$, then $P(A|B)=\frac{n(A\cap B)}{n(B)}$.

Step2: Calculate probability for on - time buses

Let $A$ be the event that a bus is owned by Jerry and $B$ be the event that a bus runs on - time. We know that $n(B) = 48$ (number of on - time buses) and $n(A\cap B)=36$ (number of on - time buses owned by Jerry). Then $P(A|B)=\frac{n(A\cap B)}{n(B)}=\frac{36}{48}=0.75 = 75.00%$.

Step3: Calculate probability for late buses

Let $A$ be the event that a bus is owned by Jerry and $B$ be the event that a bus runs late. We know that $n(B)=52$ (number of late buses) and $n(A\cap B) = 7$ (number of late buses owned by Jerry). Then $P(A|B)=\frac{n(A\cap B)}{n(B)}=\frac{7}{52}\approx0.1346\approx13.46%$.

Answer:

75.00 13.46