In a group of 30 adults, 25 are right-handed and 8 wear spectacles. The number who are right-handed and do not wear spectacles is 19.
(i) Copy and complete the following table to show the number of adults in each category.
| Wears spectacles | Does not wear spectacles | Total |
|---|
| Right-handed | | | |
| Not right-handed | | | |
| Total | | | 30 |
An adult is chosen at random from the group. Event X is ‘the adult chosen is right-handed’; event Y is ‘the adult chosen wears spectacles’.
(ii) Determine whether X and Y are independent events, justifying your answer.
Solution
(i) To complete the table, we use the given information:
- Total number of adults = 30
- Right-handed adults = 25
- Adults who wear spectacles = 8
- Right-handed and do not wear spectacles = 19
From this, we can fill in the table:
- Right-handed and wears spectacles = 25 - 19 = 6
- Not right-handed = 30 - 25 = 5
- Not right-handed and wears spectacles = 8 - 6 = 2
- Not right-handed and does not wear spectacles = 5 - 2 = 3
The completed table is:
| Wears spectacles | Does not wear spectacles | Total |
|---|
| Right-handed | 6 | 19 | 25 |
| Not right-handed | 2 | 3 | 5 |
| Total | 8 | 22 | 30 |
(ii) To determine if events X and Y are independent, we check if:
\(P(X \cap Y) = P(X) \times P(Y)\)
Calculate the probabilities:
-
\(P(X) = \frac{25}{30}\)
-
\(P(Y) = \frac{8}{30}\)
-
\(P(X \cap Y) = \frac{6}{30}\)
Check independence:
\(P(X) \times P(Y) = \frac{25}{30} \times \frac{8}{30} = \frac{200}{900} = \frac{2}{9}\)
\(P(X \cap Y) = \frac{6}{30} = \frac{1}{5}\)
Since
\(P(X \cap Y) \neq P(X) \times P(Y)\), events X and Y are not independent.
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