Question

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

• A. neither bus will be late on a particular day is $\frac{7}{10}$
• B. the bus $A$ is late given than bus $B$ is late is $\frac{9}{14}$
• C. at least one bus is late is $\frac{3}{10}$
• D. at least one bus is in time is $\frac{4}{5}$

Answer 1

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

at least one bus is late is $\frac{3}{10}$

Let $A,B$ be the events that bus $A$, bus $B$ are late respectively.
Given $P\left(A\right)=\frac{1}{5},P\left(B\right)=\frac{7}{25}\text{and}P\left(\frac{B}{A}\right)=\frac{9}{10}\Rightarrow P(A\cap B)=\frac{9}{10}\times P\left(A\right)=\frac{9}{50}$
$a\left)P\right(\overline{A}\cap \overline{B})=P\overline{(A\cup B)}=1-P(A\cup B)=1-\left[P\right(A)+P(B)-P(A\cap B\left)\right]=1-[\frac{1}{5}+\frac{7}{25}-\frac{9}{50}]=1-\frac{15}{50}=1-\frac{3}{10}=\frac{7}{10}$
$b\left)P\right(A/B)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{\left(\frac{9}{50}\right)}{\left(\frac{7}{25}\right)}=\frac{9}{14}$
$c\left)P\right(A\cup B)=(\frac{1}{5}+\frac{7}{25}-\frac{9}{50})=\frac{3}{10}$
$d\left)P\right(\overline{A}\cup \overline{B})=P(\overline{A\cap B})=1-P(A\cap B)=1-\frac{9}{50}=\frac{41}{50}$