Kate spins each spinner once - Leaving Cert Mathematics - Question b - 2021

Next, we calculate the probability that Kate lands on a red or yellow spot. Assuming there are 8 possible outcomes and 4 are either red or yellow, the calculation is:

$P(red \ or \ yellow) = \frac{4}{8} = \frac{1}{2}$

Thus, the probability that she uses a red or yellow coloured spot is ( \frac{1}{2} ).

To find the probability that Jack has his right hand and a blue spot, we need to look at the total outcomes that satisfy this condition. Assuming there are 4 outcomes for the hand and 4 for the spot, the calculation would be:

$P(right hand \ and \ blue) = P(right hand) \times P(blue) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

Therefore, the probability that the outcome is his right hand and a blue spot is ( \frac{1}{16} ).

Finally, we need to find the probability that Jack's outcome is either his right hand or a blue spot. Assuming both events are not mutually exclusive, we utilize the formula:

$P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)$

Let:

• ( P(right hand) = \frac{1}{4} )
• ( P(blue) = \frac{4}{16} = \frac{1}{4} )
• ( P(right hand \ and \ blue) = \frac{1}{16} )

Plugging in the values:

$P(right hand \ or \ blue) = \frac{1}{4} + \frac{1}{4} - \frac{1}{16} = \frac{4}{16} + \frac{4}{16} - \frac{1}{16} = \frac{7}{16}$

Thus, the probability that the outcome is his right hand or a blue spot is ( \frac{7}{16} ).