(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 o...

Answer Details

(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:

(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 \(0.152\)).