(a) A man shared his property among his…

Question 1 Report

(a) A man shared his property among his children as follows:

Child's name	Ann	Afia	Kojo	Nuno	Akom
Percentage share	5	15	10	45	25

Represent the information on a pie chart

(b) A box contains 5 red, 3 green and 4 blue identical beads. Calculate the probability th a girl takes away two red beads, one after the other, from the box.

Answer

(a) Pie chart of the property share

The shares sum to \(5+15+10+45+25 = 100\). A full circle is \(360^\circ\), so each child's sector angle is their percentage of \(360^\circ\), i.e. percentage \(\times \frac{360}{100} = \text{percentage} \times 3.6^\circ\):

Child	Percentage share (%)	Sector angle
Ann	5	\(\frac{5}{100}\times 360^\circ = 18^\circ\)
Afia	15	\(\frac{15}{100}\times 360^\circ = 54^\circ\)
Kojo	10	\(\frac{10}{100}\times 360^\circ = 36^\circ\)
Nuno	45	\(\frac{45}{100}\times 360^\circ = 162^\circ\)
Akom	25	\(\frac{25}{100}\times 360^\circ = 90^\circ\)
Total	100	\(360^\circ\)

Check: \(18^\circ+54^\circ+36^\circ+162^\circ+90^\circ = 360^\circ\). Drawing a circle and marking each sector with a protractor gives:

Pie chart of the property share: Ann 18°, Afia 54°, Kojo 36°, Nuno 162°, Akom 90° (total 360°).

(b) Probability of taking two red beads, one after the other

The box holds \(5 + 3 + 4 = 12\) identical beads, of which 5 are red. The two beads are taken one after the other without replacement.

For the first draw, 5 of the 12 beads are red:

\[P(\text{1st red}) = \frac{5}{12}.\]

After one red bead is removed, 4 red beads remain out of 11 beads in total:

\[P(\text{2nd red}\mid\text{1st red}) = \frac{4}{11}.\]

Since both events must happen, multiply the probabilities:

\[P(\text{two red}) = \frac{5}{12}\times\frac{4}{11} = \frac{20}{132} = \frac{5}{33}\approx 0.1515.\]

Therefore the probability of taking two red beads one after the other is \(\dfrac{5}{33}\) (about \(0.152\)).

Read lesson note on Probability (WAEC)

Read lesson note on Surds (radicals) (WAEC)

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