# GCSE Probability Question

• Oct 22nd 2009, 01:25 PM
GCSE Probability Question
Hi all, I'm stuck on this question from a mock paper. Any help you can give would be greatly appreciated. Thanks.

Paula is going to France for a holiday. She can travel by ferry, plane or rail. The probability that she goes by ferry is 0.2 and the probability that she goes by plane is 0.45

If Paula goes by ferry, the probability that she stays a night in Paris is 0.8, but if she goes by plane, the probability that she stays a night in Paris is 0.3

If she goes by rail, the probability that she stays a night in Paris is 0.5

What is the probability that she stays in Paris?
• Oct 22nd 2009, 02:58 PM
Soroban

Quote:

Paula is going to France for a holiday. She can travel by ferry, plane or rail.
The probability that she goes by ferry is 0.2 and the probability that she goes by plane is 0.45

If Paula goes by ferry, the probability that she stays a night in Paris is 0.8
If she goes by plane, the probability that she stays a night in Paris is 0.3
If she goes by rail, the probability that she stays a night in Paris is 0.5

What is the probability that she stays in Paris?

Given: .$\displaystyle \begin{array}{ccc} P(\text{ferry}) &=& 0.20 \\ P(\text{plane}) &=& 0.45 \end{array}$

Hence: . $\displaystyle P(\text{rail}) \;\;\;=\;\;0.35$

Also: .$\displaystyle \begin{array}{cccc}P(\text{Paris}\:|\:\text{ferry} ) &=& 0.8 & [1] \\ P(\text{Paris}\:|\:\text{plane}) &=& 0.3 & [2] \\ P(\text{Paris}\:|\:\text{rail}) &=& 0.5 & [3] \end{array}$

From [1]: .$\displaystyle \frac{P(\text{Paris}\:\wedge\text{ferry})}{P(\text {ferry})} \;=\;0.8 \quad\Rightarrow\quad \frac{P(\text{Paris}\:\wedge\:\text{ferry})}{0.20} \:=\:0.8 \quad\Rightarrow$ . $\displaystyle P(\text{Paris} \:\wedge\:\text{ferry}) \:=\:0.16$

From [2]: .$\displaystyle \frac{P(\text{Paris}\:\wedge\text{plane})}{P(\text {plane})} \;=\;0.3 \quad\Rightarrow\quad \frac{P(\text{Paris}\:\wedge\:\text{plane})}{0.45} \:=\:0.3 \quad\Rightarrow$ . $\displaystyle P(\text{Paris} \:\wedge\:\text{plane}) \:=\:0.135$

From [3]: .$\displaystyle \frac{P(\text{Paris}\:\wedge\text{rail})}{P(\text{ rail})} \;=\;0.5 \quad\Rightarrow\quad \frac{P(\text{Paris}\:\wedge\:\text{rail})}{0.35} \:=\:0.5 \quad\Rightarrow$ . $\displaystyle P(\text{Paris} \:\wedge\:\text{rail}) \:=\:0.175$

$\displaystyle \text{Therefore: }\;P(\text{Paris}) \;=\;\underbrace{P(\text{Paris}\:\wedge\:\text{fer ry})}_{0.16} + \underbrace{P(\text{Paris}\:\wedge\:\text{plane})} _{0.135} +$ $\displaystyle \underbrace{P(\text{Paris}\:\wedge\:\text{rail})}_ {0.175} \;\;=\;\;0.47$