WAEC Further Mathematics Past Questions & Answers - Page 122

The probability of picking a black ball = \(\frac{6}{6 + 14} = \frac{6}{20}\)

Probability of a white ball without replacement = \(\frac{14}{19}\)

Probability of a black ball and then white without replacement = \(\frac{6}{20} \times \frac{14}{19} = \frac{21}{95} = 0.22\)

Let the probability of getting a head = p = \(\frac{1}{2}\) and that of tail = q = \(\frac{1}{2}\)

\((p + q)^{3} = p^{3} + 3p^{2}q + 3pq^{2} + q^{3}\)

In the equation above, \(p^{3}\) and \(q^{3}\) are the probabilities of 3 heads and 3 tails respectively 

 while, \(p^{2}q\) and \(pq^{2}\) are the probabilities of 2 heads and one tail and 2 tails and one head respectively.

Probability of exactly 2 heads = \(3p^{2}q = 3(\frac{1}{2})^{2}(\frac{1}{2})\)

= \(\frac{3}{8}\)

\(Variance (\sigma^{2}) = \frac{\sum (x - \mu)^2}{n}\)

The mean \((\mu)\) of the data = \(\frac{11 + 12 + 13 + 14 + 15}{5} = \frac{65}{5} = 13\)

\(\sigma^{2} = \frac{10}{5} = 2\)

The Spearman's correlation coefficient \(\rho\) is given as:

\(\rho = 1 - \frac{6\sum d^{2}}{n(n^{2} - 1)}\)

= \(\rho = 1 - \frac{6 \times 20}{10(10^{2} - 1)}\)

= \(1 - \frac{120}{990} = \frac{870}{990}\)

= \(0.879\)

\(x^{2}\sqrt{x} \equiv x^{2}. x^{\frac{1}{2}} = x^{\frac{5}{2}}\)

\(\implies \frac{\mathrm d y}{\mathrm d x} = x^{\frac{5}{2}}\)

\(y = \int x^{\frac{5}{2}} \mathrm d x\)

= \(\frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} + c\)

= \(\frac{2x^{\frac{7}{2}}}{7} + c\)