Flags from four African countries and three European countries were displayed in a row during the 2021 Olympics - NSC Mathematics - Question 10 - 2022 - Paper 1

When calculating the probability that the African flags are displayed next to each other, we treat the group of African flags as one single unit. This means we have 4 African flags and 3 European flags, making it effectively 4 + 1 = 5 units to arrange.

The total arrangements of the units is ( 5! ), and the arrangements of the African flags within their unit is ( 4! ).

Therefore, the total arrangements with respect to the African flags together is:

$$5! \times 4! = 120 \times 24 = 2880$$

Thus, the probability is then calculated as:

$$P(African \; together) = \frac{2880}{5040} = \frac{4}{7} \approx 0.5714$$

Using the formula for independent events:

$$P(A \; or \; B) = P(A) + P(B) - P(A \; and \; B)$$

For independent events, ( P(A ; and ; B) = P(A) \times P(B) ). Therefore, substituting in:

$$0.88 = 0.4 + P(B) - (0.4 \times P(B))$$

Letting ( P(B) = x ), we have:

$$0.88 = 0.4 + x - 0.4x \Rightarrow 0.88 = 0.4 + x(1 - 0.4) \Rightarrow 0.88 - 0.4 = 0.6x \Rightarrow 0.48 = 0.6x \Rightarrow x = \frac{0.48}{0.6} = 0.8$$

Thus, ( P(B) = 0.8 ).

Given that the probability of the first passenger choosing a meat sandwich is:

$$P(M) = \frac{18}{85}$$

The total number of choices is 120, and since the choices are either meat or cheese:

$$P(C) = 1 - P(M) = 1 - \frac{18}{85} = \frac{67}{85}$$

Thus, the probability that the first passenger will choose a cheese sandwich is:

$$P(C) = \frac{67}{85} \approx 0.7882$$