A box contains 5 red, 7 blue and 4 green identical... | WAEC SSCE - Further Mathematics - 2021 (Essay)

Question 1 Report

A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement.

Calculate the probability of picking:

(a) same color of bulbs; (6) different color of bulbs (c) at least one red bulb.

Answer

The box holds \(5\) red, \(7\) blue and \(4\) green bulbs, giving \(5+7+4=16\) bulbs. Two bulbs are drawn without replacement, so the number of equally likely selections is

\[\binom{16}{2}=\frac{16\times 15}{2}=120.\]

(a) Same colour

Count the same-colour pairs for each colour and add:

Both red: \(\binom{5}{2}=10\)
Both blue: \(\binom{7}{2}=21\)
Both green: \(\binom{4}{2}=6\)

Favourable outcomes \(=10+21+6=37\), so

\[P(\text{same colour})=\frac{37}{120}.\]

(b) Different colours

Every draw is either the two bulbs matching or not, so "different colours" is the complement of "same colour":

\[P(\text{different})=1-\frac{37}{120}=\frac{83}{120}.\]

(c) At least one red

Use the complement "no red bulb". There are \(16-5=11\) non-red bulbs, giving \(\binom{11}{2}=\frac{11\times 10}{2}=55\) all-non-red pairs. Hence

\[P(\text{no red})=\frac{55}{120}=\frac{11}{24},\qquad P(\text{at least one red})=1-\frac{11}{24}=\frac{13}{24}.\]

Read lesson note on Probability (WAEC)

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