Chapter 11.4, Problem 25PPSE

Solution

To identify:The probability that It was Bus 65. You can take Bus 65 or Bus 79. You take the first bus that arrives. The probability that Bus 65 arrives first is 75%. There is a 10% chance that Bus 65 picks up passengers along the way. There is to chance that Bus 79 picks up passengers. Your bus picked up passengers.

The probability that your bus was Bus 65 is 0.6667 or 66.67%

Given information:

Bus 65 or Bus 79. You take the first bus that arrives. The probability that Bus 65 arrives first is 75%. There is a 10% chance that Bus 65 picks up passengers along the way. There is to chance that Bus 79 picks up passengers.

Explanation:

Consider the given information.

Let B65 shows the event that Bus 65 arrives first and B79 shows the event that Bus 79 arrives first. And let P shows the event that bus picks up the passenger and NP shows the event that bus does not pick up the passenger. There is a 75% chance that bus 65 arrives first. So $P\left(B65\right)\text{ =}0.75andP\left(B79\right)=\mathrm{0.25.}$ There is only 40% chance that Bus 65 picks up the = passenger so $P\left(P\text{/}B65\right)=0.40$ and there is only 60% chance that Bus 79 picks up the passenger so $P\left(P\text{/B}79\right)\text{= }\mathrm{0.60.}$ Therefore the tree diagram will be

You have to find the probability that the bus was Bus 65 given that bus picked up the passengers. So the required probability $P\left(B65/P\right)$ will be

  $P\left(B65\text{ /}P\right)=\frac{P\left(B65andP\right)}{P\left(P\right)}$

  $\begin{array}{l}=\frac{P\left(B65andP\right)}{P\left(B65andP\right)+P\left(B79andP\right)}\\ =\frac{P\left(B65\right)\cdot P\left(P/B65\right)}{P\left(B65\right)P\left(P/B65\right)+P\left(B79\right)\cdot P\left(P/B79\right)}\\ =\frac{0.75\cdot 0.40}{0.75\cdot 0.40+0.25\cdot 0.60}\\ =\frac{0.3}{0.3+0.15}\\ 0.6667\end{array}$

So, the probability that your bus was Bus 65 is 0.6667 or 66.67%