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June 2003 p6 q6
2649
The people living in 3 houses are classified as children (C), parents (P) or grandparents (G). The numbers living in each house are shown in the table below.
House number 1
House number 2
House number 3
4C, 1P, 2G
2C, 2P, 3G
1C, 1G
All the people in all 3 houses meet for a party. One person at the party is chosen at random. Calculate the probability of choosing a grandparent. [2]
A house is chosen at random. Then a person in that house is chosen at random. Using a tree diagram, or otherwise, calculate the probability that the person chosen is a grandparent. [3]
Given that the person chosen by the method in part (ii) is a grandparent, calculate the probability that there is also a parent living in the house. [4]
Solution
(i) To find the probability of choosing a grandparent, calculate the total number of grandparents and divide by the total number of people. There are 6 grandparents (2 in House 1, 3 in House 2, 1 in House 3) and 16 people in total (7 in House 1, 7 in House 2, 2 in House 3). Thus, the probability is \(\frac{6}{16} = \frac{3}{8}\).
(ii) To find the probability that a person chosen is a grandparent, calculate the probability for each house and sum them. The probability for House 1 is \(\frac{1}{3} \times \frac{2}{7}\), for House 2 is \(\frac{1}{3} \times \frac{3}{7}\), and for House 3 is \(\frac{1}{3} \times \frac{1}{2}\). Therefore, the total probability is \(\frac{1}{3} \times \frac{2}{7} + \frac{1}{3} \times \frac{3}{7} + \frac{1}{3} \times \frac{1}{2} = \frac{17}{42}\).
(iii) Given that the person chosen is a grandparent, calculate the probability that there is also a parent in the house. For House 1, the probability is \(\frac{2}{21}\), and for House 2, it is \(\frac{3}{21}\). Thus, the probability is \(\frac{2}{21} + \frac{3}{21} = \frac{5}{21}\). Divide by the probability from part (ii), \(\frac{17}{42}\), to get \(\frac{5}{21} \div \frac{17}{42} = \frac{10}{17}\).
