(a) Use logarithm tables to evaluate \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\).
(b) The first 5 students to arrive in a school on a Monday morning were 2 boys and 3 girls. Of these, two were chosen at random for an assignment. Find the probability that :
(i) both were boys ; (ii) the two were of different sexes.
Explanation
(a) \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\)
| No | Log |
| 15.05 | \(1.1775 = 1.1775\) |
| \(\sqrt{0.00695}\) | \(\bar{3}.8420 \div 2 = \bar{2}.9210\) |
| = 0.0985 | |
| \(6.95 \times 10^{2}\) | - 2.8420 |
| Antilog = 0.001805 | = \(\bar{3}.2565\) |
\(\therefore \frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}} \approxeq 0.00181\) (3 sig. figs)
(b) No of boys = 2, No of girls = 3
\(\therefore\) Total students = 5
(i) P(both are boys) = \(\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\)
(ii) P(both are of different sexes) = P(first a boy, then a girl) or P(first a girl, then a boy)
= \(\frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{2}{4} \)
= \(\frac{6}{20} + \frac{6}{20}\)
= \(\frac{12}{20} = \frac{3}{5}\)