Question

Two buses $A$ and $B$ are scheduled to arrive at a town central bus station at noon. The probability that bus $A$ and bus $B$ will be late are $\frac{1}{5}$ and $\frac{7}{25}$ respectively. The probability that the bus $B$ is late given that bus $A$ is late is $\frac{9}{10}.$ Then

• A. the probability that neither of the two bus is late on a particular day is $\frac{7}{10}$
• B. the probability that bus $A$ is late given that bus $B$ is late is $\frac{9}{14}$
• C. the probability that at least one bus is late is $\frac{3}{10}$
• D. the probability that at least one bus is in time is $\frac{41}{50}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A the probability that neither of the two bus is late on a particular day is $\frac{7}{10}$
B the probability that bus $A$ is late given that bus $B$ is late is $\frac{9}{14}$
C the probability that at least one bus is late is $\frac{3}{10}$
D the probability that at least one bus is in time is $\frac{41}{50}$

$P\left(A\right):$ Probability that bus $A$ will be late.
$P\left(B\right):$ Probability that bus $B$ will be late.

$P\left(A\right)=\frac{1}{5},P\left(B\right)=\frac{7}{25},P\left(B|A\right)=\frac{9}{10}$​
$P(\overline{A}\cap \overline{B})=1-P(A\cup B)$
$=1-\left[P\right(A)+P(B)-P(A\cap B\left)\right]$
$=1-\left[P\right(A)+P(B)-P(A\left)P\right(B|A\left)\right]$
$=1-[\frac{1}{5}+\frac{7}{25}-\frac{1}{5}\times \frac{9}{10}]$
$=\frac{7}{10}$
Therefore, the probability that neither of the two bus is late on a particular day is $\frac{7}{10}.$

$P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}$
$=\frac{P\left(A\right)P\left(B|A\right)}{P\left(B\right)}$
$=\frac{\frac{1}{5}\times \frac{9}{10}}{\frac{7}{25}}$
$=\frac{9}{14}$
Therefore, the probability that bus $A$ is late given that bus $B$ is late is $\frac{9}{14}.$

Probability that at least one bus is late
$=1-P\left(\text{none of the bus is late}\right)$
$=1-\frac{7}{10}$
$=\frac{3}{10}$

$P(A^{\prime }\cup B^{\prime })=1-P(A\cap B)$
$=1-P\left(A\right)P\left(B|A\right)$
$=1-\frac{1}{5}\times \frac{9}{10}$
$=\frac{41}{50}$
Therefore, the probability that at least one bus is in time is $\frac{41}{50}.$