Theory question 106
Logarithms(a) Use logarithm tables to evaluate (15.05 × √0.00695)/(6.95 × 102), correct to 3 significant figures.
(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.
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(a) log 15.05 = 1.1775; log 0.00695 = 3̄.8420, half of it = 2̄.9210 (write 4̄ + 1.8420, divide by 2); log 695 = 2.8420.
Numerator: 1.1775 + 2̄.9210 = 0.0985. Subtract 2.8420: 3̄.2565.
Antilog = 0.001806 ≈ 0.00181.
(b) Two are chosen without replacement.
(i) P(both boys) = 2/5 × 1/4 = 1/10.
(ii) P(different sexes) = 2/5 × 3/4 + 3/5 × 2/4 = 6/20 + 6/20 = 3/5.
Two different students are chosen, so the second pick is from 4, not 5.