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Two buses A and B are scheduled to arrive at a town central bus station at noon. The probability that bus A will be late is 1/5. The probability that bus B will be late is 7/25. The probability that the bus B is late given that bus A is late is 9/10. Then the probabilities:
 (i) Neither bus will be late on a particular day and
(ii) Bus A is late given that bus B is late, are respectively.
(A). 2/25 and 12/28.
(B). 18/25 and 22/28.
(C). 7/10 and 18/28.
(D). 12/25 and 2/28.

Answer Verified Verified
Hint: We will be using the concepts of probability to solve the problem. We will be using the concepts of compound probability to further simplify the problem.

Complete step-by-step solution -
Now we have been given that two buses A and B are scheduled to arrive at a town central bus station at noon.
The probability that the bus A is late
$P\left( A \right)=\dfrac{1}{5}$ ……………………. (1)
The probability that bus B is late
$P\left( B \right)=\dfrac{7}{25}$ …………………. (2)
Probability that B is late given that A is late
$\dfrac{9}{10}=P\left( \dfrac{B}{A} \right)$ ……………….. (3)
Now, we have to find the probability that wither bus will be late on a particular day is $P\left( \overline{A}\cap \overline{B} \right)$ .
Now, we know that \[P\left( \dfrac{B}{A} \right)=\dfrac{P\left( B\cap A \right)}{P\left( A \right)}\] .
Now, we will substitute the value of $P\left( {B}/{A}\; \right)$ and $P\left( A \right)$ from (1) and (3) so that we have
$\dfrac{1}{5}\times \dfrac{9}{10}=P\left( B\cap A \right)$ .
$P\left( B\cap A \right)=\dfrac{9}{50}$ ……………… (4)
Now we know that $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ ,
So, we will substitute the value of $P\left( A\cap B \right)$ , $P\left( A \right)$ and $P\left( B \right)$.
$\begin{align}
  & P\left( A\cup B \right)=\dfrac{1}{5}+\dfrac{7}{25}-\dfrac{9}{50} \\
 & =\dfrac{12}{25}-\dfrac{9}{50} \\
 & =\dfrac{24-9}{50} \\
 & =\dfrac{15}{50} \\
 & P\left( A\cup B \right)=\dfrac{3}{10} \\
\end{align}$
Now, we have to find the value of $P\left( \overline{A}\cap \overline{B} \right)$ we know that,
$P\left( \overline{A}\cap \overline{B} \right)=P\left( \overline{A\cup B} \right)$ by demerger’s law.
$P\left( \overline{A}\cap \overline{B} \right)=1-P\left( A\cup B \right)$ .
Now, we will substitute the value of $P\left( A\cup B \right)$ .therefore,
$\begin{align}
  & P\left( \overline{A}\cap \overline{B} \right)=1-\dfrac{3}{10} \\
 & =\dfrac{7}{10} \\
\end{align}$
So, the probability that neither bus will be late $=\dfrac{7}{10}$.
Now, we have to find the probability that bus A is late given that bus B is late is $P\left( \dfrac{A}{B} \right)$ .
Now, we know that $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$.
Now, we will substitute the value from (2) and (4)
\[\begin{align}
  & =\dfrac{9}{50}\times \dfrac{25}{7} \\
 & =\dfrac{9}{14} \\
 & =\dfrac{18}{28} \\
\end{align}\]
Hence the correct option is (B).

Note: To solve these types of question we must have a basic understanding of probability also it is important to remember that
$\begin{align}
  & P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)} \\
 & P\left( \overline{A} \right)=1-P\left( A \right) \\
 & P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right). \\
\end{align}$
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