(a) The probabilities that three boys pass an... | WAEC SSCE - General Mathematics - 2001 (Essay)

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(a) The probabilities that three boys pass an examination are \(\frac{2}{3}, \frac{5}{8}\) and \(\frac{3}{4}\) respectively. Find the probability that :

(i) all three boys pass ; (ii) none of the boys pass ; (iii) only two of the boys pass.

(b) A shop-keeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% of the marked price. Calculate the actual percentage profit.

Answer

(a) Pass probabilities \(\tfrac{2}{3},\tfrac{5}{8},\tfrac{3}{4}\); fail probabilities \(\tfrac{1}{3},\tfrac{3}{8},\tfrac{1}{4}\).

(i) All three pass \(=\dfrac{2}{3}\times\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{30}{96}=\dfrac{5}{16}\).

(ii) None passes \(=\dfrac{1}{3}\times\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{96}=\dfrac{1}{32}\).

(iii) Exactly two pass:

1&2 pass, 3 fails: \(\tfrac{2}{3}\cdot\tfrac{5}{8}\cdot\tfrac{1}{4}=\tfrac{10}{96}\)
1&3 pass, 2 fails: \(\tfrac{2}{3}\cdot\tfrac{3}{8}\cdot\tfrac{3}{4}=\tfrac{18}{96}\)
2&3 pass, 1 fails: \(\tfrac{1}{3}\cdot\tfrac{5}{8}\cdot\tfrac{3}{4}=\tfrac{15}{96}\)

Total \(=\dfrac{10+18+15}{96}=\dfrac{43}{96}\).

(b) Marked price \(=\text{N}36{,}000\) gives \(20\%\) profit, so cost price \(=\dfrac{36000}{1.20}=\text{N}30{,}000\).

Discount of \(5\%\): selling price \(=36000\times 0.95=\text{N}34{,}200\).

Profit \(=34200-30000=\text{N}4{,}200\); actual percentage profit \(=\dfrac{4200}{30000}\times 100=\mathbf{14\%}\).

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Answer Details

(a) Pass probabilities \(\tfrac{2}{3},\tfrac{5}{8},\tfrac{3}{4}\); fail probabilities \(\tfrac{1}{3},\tfrac{3}{8},\tfrac{1}{4}\).

(i) All three pass \(=\dfrac{2}{3}\times\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{30}{96}=\dfrac{5}{16}\).

(ii) None passes \(=\dfrac{1}{3}\times\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{96}=\dfrac{1}{32}\).

(iii) Exactly two pass:

1&2 pass, 3 fails: \(\tfrac{2}{3}\cdot\tfrac{5}{8}\cdot\tfrac{1}{4}=\tfrac{10}{96}\)
1&3 pass, 2 fails: \(\tfrac{2}{3}\cdot\tfrac{3}{8}\cdot\tfrac{3}{4}=\tfrac{18}{96}\)
2&3 pass, 1 fails: \(\tfrac{1}{3}\cdot\tfrac{5}{8}\cdot\tfrac{3}{4}=\tfrac{15}{96}\)

Total \(=\dfrac{10+18+15}{96}=\dfrac{43}{96}\).

(b) Marked price \(=\text{N}36{,}000\) gives \(20\%\) profit, so cost price \(=\dfrac{36000}{1.20}=\text{N}30{,}000\).

Discount of \(5\%\): selling price \(=36000\times 0.95=\text{N}34{,}200\).

Profit \(=34200-30000=\text{N}4{,}200\); actual percentage profit \(=\dfrac{4200}{30000}\times 100=\mathbf{14\%}\).

Read lesson note on Probability (WAEC)

Read lesson note on Financial Arithmetic (WAEC)

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(a) The probabilities that three boys pass an examination are \(\frac{2}{3}, \frac{5}{8}\) and \(\frac{3}{4}\) respectively. Find the probability that : (i)...

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(a) The probabilities that three boys pass an examination are \(\frac{2}{3}, \frac{5}{8}\) and \(\frac{3}{4}\) respectively. Find the probability that : (i)...

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