Each of 90 students at a sports day plays football, netball, both or neither. In the Venn diagram \(x\) play both and \(2x\) play neither. FN38x252x© EAGLE ...
Question 1 Report
Each of 90 students at a sports day plays football, netball, both or neither. In the Venn diagram \(x\) play both and \(2x\) play neither.
- Show that \(x = 9\). (2)
- A student is picked at random. Find the probability that this student plays netball. (1)
- A netball player is picked at random. Find the probability that she also plays football. (2)
Answer Details
The four regions of the diagram do not overlap and account for all \(90\) students, so their expressions add to \(90\). Reading the diagram, \(38\) play football only, \(x\) play both, \(25\) play netball only, and \(2x\) play neither.
(a)
\[38 + x + 25 + 2x = 90\][1]
Collecting terms, the constants give \(38 + 25 = 63\) and the \(x\) terms give \(x + 2x = 3x\):
\[63 + 3x = 90 \implies 3x = 27 \implies x = 9 \text{ as required.}\][1]
(b) The netball set covers two regions, netball only and the overlap:
\[25 + 9 = 34 \text{ netball players}\] \[P(\text{netball}) = \frac{34}{90} = \frac{17}{45}\][1]
(c) This is a conditional probability. The student is chosen from the netball players only, so the denominator is \(34\) rather than \(90\). Of those \(34\), the \(9\) in the overlap also play football. [1]
\[P(\text{football given netball}) = \frac{9}{34}\][1]
The change of denominator between parts (b) and (c) is the point being tested. In part (b) the student comes from the whole year group; in part (c) the pool has already been narrowed to netball players. Using \(\frac{9}{90}\) in part (c) would throw away the information the question supplies.
Substituting \(x = 9\) checks the diagram: the regions hold \(38\), \(9\), \(25\) and \(18\) students, totalling \(90\).
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