(a) Without using Mathematical tables or calculators, simplify:
\(3\frac{4}{9} \div (5\frac{1}{3} – 2\frac{3}{4}) + 5\frac{9}{10}\)
(b) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. Find the probability that the sum of the two numbers is greater than 3 and less than 7.
Explanation
(a) \(3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}\)
\(\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4}) + \frac{59}{10}\)
\(\frac{31}{9} \div (\frac{64 - 33}{12}) + \frac{59}{10}\)
\((\frac{31}{9} \div \frac{31}{12}) + \frac{59}{10}\)
\((\frac{31}{9} \times \frac{12}{31}) + \frac{59}{10}\)
\(\frac{12}{9} + \frac{59}{10} = \frac{120 + 531}{90}\)
\(\frac{651}{90} = \frac{217}{30}\).
(b)
| + | 1 | 3 | 5 |
| 2 | 3 | 5 | 7 |
| 3 | 4 | 6 | 8 |
| 4 | 5 | 7 | 9 |
Let E be the event of the sum being greater than 3 and less than 7 and S be the total sample space.
n(E) = 4; and n(S) = 9.
P(E) = \(\frac{n(E)}{n(S)} = \frac{4}{9}\)