(a) Without using Mathematical tables or calculators, simplify:
(3frac{4}{9} div (5frac{1}{3} – 2frac{3}{4}) + 5frac{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) (3frac{4}{9} div (5frac{1}{3} – 2frac{3}{4}) + 5frac{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})