a) An experiment consists of throwing two fair, standard, six-sided dice and recording the sum of the two numbers thrown - Leaving Cert Mathematics - Question 1 - 2018

The favorable outcomes for a total of 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) which gives 6 outcomes. For a total of 11, the outcomes are: (5,6), (6,5) which gives 2 outcomes. Thus, the total favorable outcomes for either 7 or 11 is:

$6 + 2 = 8$

The total possible outcomes when throwing two dice is 36.

Therefore, the probability is:

$P(7 ext{ or } 11) = \frac{8}{36} = \frac{2}{9}$

The prime outcomes possible with two dice are 2, 3, 5, 7, 11. The favorable combinations are:

• Total 2: (1,1) – 1 outcome
• Total 3: (1,2), (2,1) – 2 outcomes
• Total 5: (1,4), (2,3), (3,2), (4,1) – 4 outcomes
• Total 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) – 6 outcomes
• Total 11: (5,6), (6,5) – 2 outcomes

Adding these gives:

$1 + 2 + 4 + 6 + 2 = 15$

Thus, the probability is:

$P(prime) = \frac{15}{36} = \frac{5}{12}$

Total number of cars = 30. Total Ford cars = 18, out of which 7 are red. Total Renault cars = 12, out of which 4 are red.

Thus, the number of non-red cars:

• Non-red Ford = 18 - 7 = 11
• Non-red Renault = 12 - 4 = 8

Total non-red cars = 11 + 8 = 19.

Therefore, the probability that the car chosen is a Ford or non-red is:

$P(Ford ext{ or not red}) = \frac{18 + 19}{30} = \frac{37}{30}$