Each passenger on a certain Banana Airways flight chose exactly one beverage from tea, coffee or fruit juice - NSC Mathematics - Question 10 - 2016 - Paper 1

The probability of selecting a male passenger is given by the ratio of the number of male passengers to the total number of passengers:

$P(M) = \frac{60}{160} = \frac{3}{8} = 0.375$

Using the formula for independence,

$P(Male) \times P(Coffee) = P(Male \text{ and } Coffee)$

Let $b$ be the number of male passengers who chose coffee:

$P(Male) = \frac{60}{160}, P(Coffee) = \frac{80}{160}$

Thus,

$\frac{60}{160} \times \frac{80}{160} = \frac{b}{160}$

Solving for $b$:

$b = \frac{60 \times 80}{160} = 30$

The number of arrangements for 6 people in 6 seats is calculated as:

$6! = 720$

We can treat Xoliswa and Anees as a single entity or block. Thus, we have 5 entities to arrange:

• (XA), 4 other passengers

The arrangements are:

$5! = 120$

However, within the block (XA), Xoliswa and Anees can switch places, giving:

$5! \times 2 = 240$

If there are 6 seats, Mary has 2 options (either end of the row). The total arrangements of the other 5 passengers is:

$5! = 120$

Thus, the probability is:

$P(Mary ext{ at end}) = \frac{2 \times 120}{720} = \frac{1}{6}$