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:

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

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

= $1 - \frac{120}{990} = \frac{870}{990}$

= $0.879$

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

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

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

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

= $\frac{2x^{\frac{7}{2}}}{7} + c$