Question

2.1. on a flight, passengers could choose between a vegetarian snack and chicken snack. the snack selected by passengers were recorded. the results are shown in the table below.
was the choice of snack on this flight independent of gender? motivate your answer with the necessary calculations.
2.2.1. for any two events, a and b, it is given that ( p(a and b)=0.12 ). ( p(a or b)=0.83 ) and ( p(b)=4p(a) ).
2.2.2. calculate ( p(b) )
2.2.3. calculate ( p(nota) ).

Explanation:

Step1: Use the formula for probability of independent events

If two events \(A\) (snack choice) and \(B\) (gender) are independent, then \(P(A\cap B)=P(A)\times P(B)\). Let's check for vegetarian snack and male gender. \(P(\text{Vegetarian})=\frac{32}{150}\), \(P(\text{Male})=\frac{67}{150}\), \(P(\text{Vegetarian}\cap\text{Male})=\frac{12}{150}\). Calculate \(P(\text{Vegetarian})\times P(\text{Male})=\frac{32\times67}{150\times150}=\frac{2144}{22500}\approx0.095\), and \(\frac{12}{150} = 0.08\). Since \(0.08
eq0.095\), the choice of snack is not independent of gender.

Step2: Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) for 2.2.1

Let \(P(A) = x\), then \(P(B)=4x\). Substitute into \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we get \(0.83=x + 4x-0.12\).

Step3: Solve the equation for \(x\) (2.2.2)

Simplify \(0.83 = 5x-0.12\), then \(5x=0.83 + 0.12=0.95\), \(x=\frac{0.95}{5}=0.19\). So \(P(B)=4x = 4\times0.19 = 0.76\).

Step4: Calculate \(P(\text{not}A)\) (2.2.3)

Since \(P(A)=0.19\), \(P(\text{not}A)=1 - P(A)=1-0.19 = 0.81\).

Answer:

2.1: The choice of snack is not independent of gender.
2.2.1: Equation \(0.83=x + 4x-0.12\) is set up.
2.2.2: \(P(B)=0.76\)
2.2.3: \(P(\text{not}A)=0.81\)