Question

# 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 $\dfrac{1}{5}$. The probability that bus B will be late is $\dfrac{7}{{25}}$. The probability that the bus B is late given that bus A is late is $\dfrac{9}{{10}}$. 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). $\dfrac{2}{{25}}$ and $\dfrac{{12}}{{28}}$(b). $\dfrac{{18}}{{25}}$ and $\dfrac{{22}}{{28}}$(c). $\dfrac{7}{{10}}$ and $\dfrac{{18}}{{28}}$(d). $\dfrac{{12}}{{25}}$ and $\dfrac{2}{{28}}$

Hint: Use the formula for conditional probability, which is $P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}}$ to find the probability for the required events. This is based on conditional probability.

Let A be the event that the bus A is late. Then, we have:
$P(A) = \dfrac{1}{5}...........(1)$
Let B be the event that the bus B is late. Then, we have:
$P(B) = \dfrac{7}{{25}}...........(2)$
Condition probability for two events A and B, in which the probability of event A given event B occurs is given as follows:
$P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}}.........(3)$
It is given that the probability that the bus B is late given that bus A is late is $\dfrac{9}{{10}}$. Hence, by formula (3), we have:
$P(B|A) = \dfrac{{P(A \cap B)}}{{P(A)}}$
From equation (1), we have:
$\dfrac{9}{{10}} = \dfrac{{P(A \cap B)}}{{\dfrac{1}{5}}}$
Solving for $P(A \cap B)$, we have:
$P(A \cap B) = \dfrac{9}{{10}}.\dfrac{1}{5}$
$P(A \cap B) = \dfrac{9}{{50}}...........(4)$
We need to find the probability that neither of the buses is late given by $P(\bar A \cap \bar B)$.
Using, De Morgan’s Law, we have:
$P(\bar A \cap \bar B) = P(\overline {A \cup B} )$
$P(\bar A \cap \bar B) = 1 - P(A \cup B).........(5)$
Hence, we need to find $P(A \cup B)$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, hence from equations (1), (2) and (3), we have:
$P(A \cup B) = \dfrac{1}{5} + \dfrac{7}{{25}} - \dfrac{9}{{50}}$
Simplifying, we get:
$P(A \cup B) = \dfrac{{10 + 14 - 9}}{{50}}$
$P(A \cup B) = \dfrac{{15}}{{50}}$
$P(A \cup B) = \dfrac{3}{{10}}$
Hence, substituting the value in equation (6), we have:
$P(\bar A \cap \bar B) = 1 - \dfrac{3}{{10}}$
$P(\bar A \cap \bar B) = \dfrac{7}{{10}}$
Hence, the probability that neither of the buses is late is $\dfrac{7}{{10}}$.
Next, we need to find the probability that bus A is late given that bus B is late. Using conditional probability formula in equation (3), we have:
$P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}}$
$P(A|B) = \dfrac{{\dfrac{9}{{50}}}}{{\dfrac{7}{{25}}}}$
Simplifying we have:
$P(A|B) = \dfrac{9}{{14}}$
Multiplying numerator and denominator by 2, we have:
$P(A|B) = \dfrac{{18}}{{28}}$
Hence, the probability that bus A arrives late given that bus B is late is $\dfrac{{18}}{{28}}$.
Hence, the correct answer is option (c).

Note: You can also find the probability that bus A is not late and the bus B is not late and then proceed to solve for the probability that both the buses are not late. In this type of question when conditional probability is asked like probability of event “A” when event “B” occurs we use the formula $P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}}$ .