Question

a family from the survey is selected at random. match the probability to each event.
number of children
car
bus
total
p(car|1 - 2 children)
p(3 + children|bus)
p(1 - 2 children|car)·p(3 + children|car)
p(bus|1 - 2 children)·p(bus|3 + children)
0.203
0.284
0.563
0.249
0.624
0.662

Explanation:

Step1: Calculate \(P(\text{Car}|1 - 2\text{ Children})\)

The formula for conditional probability is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here, \(A\) is "Car" and \(B\) is "1 - 2 Children". \(P(\text{Car}\cap1 - 2\text{ Children})=\frac{63}{175}\), \(P(1 - 2\text{ Children})=\frac{101}{175}\). So \(P(\text{Car}|1 - 2\text{ Children})=\frac{63}{101}\approx0.624\).

Step2: Calculate \(P(3+\text{ Children}|\text{Bus})\)

Using the conditional - probability formula \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here, \(A\) is "3 + Children" and \(B\) is "Bus". \(P(3+\text{ Children}\cap\text{Bus})=\frac{49}{175}\), \(P(\text{Bus})=\frac{87}{175}\). So \(P(3+\text{ Children}|\text{Bus})=\frac{49}{87}\approx0.563\).

Step3: Calculate \(P(1 - 2\text{ Children}|\text{Car})\cdot P(3+\text{ Children}|\text{Car})\)

First, \(P(1 - 2\text{ Children}|\text{Car})=\frac{63}{88}\), \(P(3+\text{ Children}|\text{Car})=\frac{25}{88}\). Then \(P(1 - 2\text{ Children}|\text{Car})\cdot P(3+\text{ Children}|\text{Car})=\frac{63\times25}{88\times88}=\frac{1575}{7744}\approx0.203\).

Step4: Calculate \(P(\text{Bus}|1 - 2\text{ Children})\cdot P(\text{Bus}|3+\text{ Children})\)

\(P(\text{Bus}|1 - 2\text{ Children})=\frac{38}{101}\), \(P(\text{Bus}|3+\text{ Children})=\frac{49}{74}\). Then \(P(\text{Bus}|1 - 2\text{ Children})\cdot P(\text{Bus}|3+\text{ Children})=\frac{38\times49}{101\times74}=\frac{1862}{7474}\approx0.249\).

Answer:

From left - to - right: \(0.249\), \(0.203\), \(0.563\), \(0.624\)