A car is selected at random - Leaving Cert Mathematics - Question a - 2014

To determine the probability that the car is either black or red, we again use the probability formula:

$P(Black \text{ or } Red) = \frac{\text{Number of black cars + Number of red cars}}{\text{Total number of cars}}$

Thus,

$P(Black \text{ or } Red) = \frac{5 + 9}{24} = \frac{14}{24} = \frac{7}{12}$

For this part, we need to use conditional probability and the total probability formula:

$P(First\ Silver \text{ and } Second\ Black) = P(Silver) \times P(Black | Silver)$

The probability of selecting a silver car first is:

$P(Silver) = \frac{10}{24}$

After selecting a silver car, there are now 23 cars left, including 5 black cars:

$P(Black | Silver) = \frac{5}{23}$

Therefore,

$P(First\ Silver \text{ and } Second\ Black) = \frac{10}{24} \times \frac{5}{23} = \frac{50}{552} = \frac{25}{276}$

To find this probability, we can consider two scenarios:

1. The first car is red and the second is black.
2. The first car is black and the second is red.

Now, let's calculate:

1. For the first condition: $P(Red \text{ then } Black) = P(Red) \times P(Black | Red)$ $P(Red) = \frac{9}{24}$ After selecting a red car, there are now 23 cars left, including 5 black cars: $P(Black | Red) = \frac{5}{23}$ Thus, $P(Red \text{ then } Black) = \frac{9}{24} \times \frac{5}{23} = \frac{45}{552}$

2. For the second condition: $P(Black \text{ then } Red) = P(Black) \times P(Red | Black)$ $P(Black) = \frac{5}{24}$ After selecting a black car, there are now 23 cars left, including 9 red cars: $P(Red | Black) = \frac{9}{23}$ Thus, $P(Black \text{ then } Red) = \frac{5}{24} \times \frac{9}{23} = \frac{45}{552}$

Finally, we combine both scenarios:

$P(Red \text{ and } Black) = P(Red \text{ then } Black) + P(Black \text{ then } Red = \frac{45 + 45}{552} = \frac{90}{552} = \frac{15}{92}$

To find the probability of selecting a red car or a diesel car, we can first find the total number of red cars and the number of diesel cars:

• Total red cars = 9
• Total diesel cars = 3 (from black) + 2 (from red) + 4 (from silver) = 9

Now, since these categories overlap, we need to use the formula for the union of two events:

$P(Red \text{ or } Diesel) = P(Red) + P(Diesel) - P(Red \text{ and } Diesel)$

First, we find:

$P(Red) = \frac{9}{24}$ $P(Diesel) = \frac{9}{24}$

Considering that among the 9 red cars, none were diesel:

Thus, the total probability becomes:

$P(Red \text{ or } Diesel) = \frac{9}{24} + \frac{9}{24} - 0 = \frac{18}{24} = \frac{2}{3}$