A game is played using the two spinners shown below - Junior Cycle Mathematics - Question 3 - 2014

There are 3 outcomes that contain the letter E: (2, E), (4, E), and (6, E).

The probability that the outcome contains the letter E is given by the formula:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{3}{18} = \frac{1}{6}$

The probability that the outcome contains the number 6 is:

$P(6) = \frac{\text{Number of favorable outcomes (that contain 6)}}{\text{Total outcomes}} = \frac{6}{18} = \frac{1}{3}$

To find the probability that the outcome contains E, or 6, or both, we can use the principle of inclusion-exclusion:

$P(E \cup 6) = P(E) + P(6) - P(E \cap 6)$

Since the outcomes (6, E) is counted in both P(E) and P(6), we adjust accordingly. Thus:

$P(E \cup 6) = \frac{1}{6} + \frac{1}{3} - 0 = \frac{3}{18} + \frac{6}{18} = \frac{9}{18} = \frac{1}{2}$