In a hotel, the breakfast is a choice between yam... | WAEC SSCE - Further Mathematics - 2008 (Essay)

Question 1 Report

In a hotel, the breakfast is a choice between yam (Y) or plantain (P) or both. The Venn diagram shows the choices made by 25 guests of the hotel.

(a) Find the value of x;

(b) What is the probability that a guest chosen at random chose only one of the two?

Answer

From the Venn diagram the three disjoint regions are:

Plantain only (\(P\) region outside the overlap): \(2x+1\)
Both yam and plantain (the overlap): \(x\)
Yam only (\(Y\) region outside the overlap): \((x-2)^2\)

Every one of the 25 guests chose yam, plantain, or both, so the three regions add up to the total number of guests.

(a) Finding the value of \(x\)

\[ (2x+1) + x + (x-2)^2 = 25 \]

Expand \((x-2)^2 = x^2 - 4x + 4\):

\[ 2x + 1 + x + x^2 - 4x + 4 = 25 \]\[ x^2 - x + 5 = 25 \]\[ x^2 - x - 20 = 0 \]

Factorising:

\[ (x-5)(x+4) = 0 \]\[ x = 5 \quad \text{or} \quad x = -4 \]

A number of guests cannot be negative, so we reject \(x = -4\).

\[ \boxed{x = 5} \]

(b) Probability that a guest chose only one of the two

Substitute \(x = 5\) into each region:

\[ \text{Plantain only} = 2(5)+1 = 11 \]\[ \text{Both} = 5 \]\[ \text{Yam only} = (5-2)^2 = 9 \]

Check: \(11 + 5 + 9 = 25\), which matches the total number of guests.

Guests who chose only one of the two are those in "plantain only" or "yam only":

\[ 11 + 9 = 20 \]

Therefore the required probability is:

\[ P(\text{only one}) = \frac{20}{25} = \frac{4}{5} = 0.8 \]

Read lesson note on Sets (WAEC)

Read lesson note on Probability (WAEC)

Answer Details