John played Game A four times and tells us that he has won a total of €8 - Leaving Cert Mathematics - Question 3 - 2014

To determine which game is more successful for raising funds, we need to calculate the expected outcomes for each game:

• For Game A:

  $E(X) = 0(\frac{1}{6}) + 3(\frac{1}{6}) + 5(\frac{1}{6}) + 6(\frac{1}{6}) = \frac{14}{6} = 2.33$

• For Game B:

  $E(X) = 0(\frac{1}{6}) + 1(\frac{1}{6}) + 2(\frac{1}{6}) + 3(\frac{1}{6}) + 4(\frac{1}{6}) = \frac{10}{6} = 1.67$

Based on the calculations, Game A yields a higher expected payout, thus is more successful in raising funds for the charity.

To find the probability that the arrow stops in the €4 sector exactly twice when playing Game B six times, we can use the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• (n) = total number of trials (6 spins)
• (k) = number of successful outcomes (2 stops in €4 sector)
• (p) = probability of success (rac{1}{6})
• (1 - p) = probability of failure (rac{5}{6})

Substituting the values, we get:

$P(stops \ in \ €4 \ sector \ exactly \ twice) = \binom{6}{2} (\frac{1}{6})^2 (\frac{5}{6})^{4} = 15 \times (\frac{1}{36}) \times (\frac{625}{1296}) = 0.2$