Paul is raising money for a charity in his school - Junior Cycle Mathematics - Question 2 - 2016

To get €8 back, the sum of the numbers on the two spinners must equal 8.

From the table, we see that the pairs (3, 5), (4, 4), and (5, 3) yield a sum of 8.

Counting the outcomes:

• (3, 5)
• (4, 4)
• (5, 3)

There are 3 favorable outcomes out of a total of 25 (5x5), so the probability is:

$P(€8) = \frac{3}{25} = \frac{1}{15}$

The expected number of students getting €1 back is calculated using the probability found in part (b).
Since we found that the probability of getting €1 back is (\frac{3}{15}), we can calculate:

$E(€1) = 320 \times \frac{3}{15} = 64$

Let x be the number of students who got €8 back.
We know that:

$320 = 74 + x + 74 + 110$
Substituting the numbers:

• 320 = 74 + x + 110
• 320 - 74 - 110 = x
• 136 = x

Therefore, the number of students who got €8 back is:

• x = 17

To analyze the change in probability, we must create a new two-way table for Spinner B with outcomes from 1 to 6.

New Two-Way Table:

	Spinner B 1	Spinner B 2	Spinner B 3	Spinner B 4	Spinner B 5	Spinner B 6
Spinner A 1	2	3	4	5	6	7
Spinner A 2	3	4	5	6	7	8
Spinner A 3	4	5	6	7	8	9
Spinner A 4	5	6	7	8	9	10
Spinner A 5	6	7	8	9	10	11
Spinner A 6	7	8	9	10	11	12

After analyzing the new probabilities, it is evident that the potential combinations yielding €1 or €8 have increased, hence Paul's statement is incorrect.