(a) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. What is the probability that the sum of the two numbers will be less than ...

Answer Details

(a) One number is chosen from \(\{2,3,4\}\) and one from \(\{1,3,5\}\). There are \(3\times 3 = 9\) equally likely pairs. We need the sum \(S\) to satisfy \(3 < S < 7\), i.e. \(S \in \{4,5,6\}\).

+	1	3	5
2	3	5	7
3	4	6	8
4	5	7	9

The sums strictly between \(3\) and \(7\) are: \((2,3)=5,\;(3,1)=4,\;(3,3)=6,\;(4,1)=5\) - that is \(4\) favourable outcomes.

\[P(3 < S < 7) = \frac{4}{9}.\]

(b) \(ABCD\) is a circle. \(DAE\) and \(CBE\) are straight lines meeting at the outer point \(E\) (with \(\angle E = m\)); \(ABF\) and \(DCF\) are straight lines meeting at the outer point \(F\) (with \(\angle F = y\)). Let the four arcs be \(AB = p,\; BC = q,\; CD = r,\; DA = s\), so \(p+q+r+s = 360^\circ\).

Angle at each external point = half the difference of the intercepted arcs.

\[m = \angle E = \tfrac12(\text{arc }DC - \text{arc }AB) = \tfrac12(r-p),\qquad y = \angle F = \tfrac12(\text{arc }AD - \text{arc }BC) = \tfrac12(s-q).\]

Given \(y + m = 90^\circ\):

\[\tfrac12(r-p) + \tfrac12(s-q) = 90^\circ \;\Rightarrow\; (r+s) - (p+q) = 180^\circ.\]

Combine with \(p+q+r+s = 360^\circ\). Adding the two equations gives \(2(r+s) = 540^\circ\), so

\[r+s = 270^\circ,\qquad p+q = 90^\circ.\]

At vertex \(B\) the straight lines \(CBE\) and \(ABF\) cross, and \(x\) is the angle \(\angle ABE = \angle FBC\) (vertically opposite), which is the supplement of the interior angle \(\angle ABC\). Now \(\angle ABC\) is inscribed on arc \(ADC\):

\[\angle ABC = \tfrac12(\text{arc }AD + \text{arc }DC) = \tfrac12(s+r) = \tfrac12(270^\circ) = 135^\circ.\]

Therefore

\[x = 180^\circ - \angle ABC = 180^\circ - 135^\circ = \boxed{45^\circ}.\]