Two events A and B are such that $P(A) = \frac{3}{4}$ and $P(A \cap B) = \frac{1}{2}$ - Leaving Cert Mathematics - Question 5 - 2021

To determine if events A and B are independent, we check if:

$P(A \cap B) = P(A) \times P(B)$

First, we need to find $P(B)$ using:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substituting the known values:

$\frac{11}{12} = \frac{3}{4} + P(B) - \frac{1}{2}$

Calculating $P(B)$:

$\frac{11}{12} = \frac{9}{12} + P(B) - \frac{6}{12}$ $\frac{11}{12} = \frac{3}{12} + P(B)$ $P(B) = \frac{11}{12} - \frac{3}{12} = \frac{8}{12} = \frac{2}{3}$

Next, we check independence:

$P(A) \times P(B) = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$

Since:

$P(A \cap B) = \frac{1}{2},$

which equals $P(A) \times P(B)$, we conclude that events A and B are independent.

To determine if the game is fair, we need to analyze the outcomes from spinning the spinner twice.

The possible sums when the spinner (segments numbered 1 to 4) is spun twice:

• Each roll gives outcomes from 1 to 4, so the sums can range from 2 (1 + 1) to 8 (4 + 4).
• We must now find the remainders when these sums are divided by 3:
  • Sum 2: remainder 2
  • Sum 3: remainder 0
  • Sum 4: remainder 1
  • Sum 5: remainder 2
  • Sum 6: remainder 0
  • Sum 7: remainder 1
  • Sum 8: remainder 2

The outcomes for each player:

• Liam wins on remainders of 0: 3 and 6 (2 outcomes)
• Sorcha wins on remainders of 1: 4 and 7 (2 outcomes)
• Lee wins on remainders of 2: 2, 5, 8 (3 outcomes)

Total outcomes for a fair game would mean each player should have the same probability of winning. As it stands:

• Liam: 2 chances
• Sorcha: 2 chances
• Lee: 3 chances

Lee has the most chances to win. Since the probabilities are not equal, we conclude:

Conclusion: The game is not fair.