# On a certain road 20% of the vehicles are trucks, 16% are buses and the remainder are cars.

(i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses. [3]
(ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the
probability that more than 73 are cars.

## solution, P(X < 3) = P(0) + P(1) + P(2) M1

Binomial term with 11Cr p

= (0.84)11 + (0.16)(0.84)10 Ã— 11C1 +
(0.16)2
(0.84)9
Ã— 11C2
Correct expression for P(0, 1, 2) or P(0, 1, 2, 3)

= 0.1469 + 0.30782 + 0.2931
= 0.748
ii) Î¼ = 125 Ã— 0.64 = 80
Ïƒ^2= 125 Ã— 0.64 Ã— 0.36 = 28.8
P(X > 73) = 1 â€“phi(73.5-80/sqrt(28.8)
standardising = phi(1.211)
observe region, correct region (> 0.5 if mean > 73.5, if mean < 73.5)
= .887 ans

## (i) Well, if we want to find the probability that fewer than 3 out of 11 vehicles are buses, we can start by finding the probability that exactly 0 buses are selected and the probability that exactly 1 bus is selected. Then, we can sum up those probabilities.

The probability of selecting exactly 0 buses out of 11 can be calculated using the binomial distribution formula: (11 choose 0) * (0.16)^0 * (0.84)^11.

The probability of selecting exactly 1 bus out of 11 can be calculated using the binomial distribution formula: (11 choose 1) * (0.16)^1 * (0.84)^10.

Finally, we can sum up these two probabilities to get the probability of selecting fewer than 3 buses.

(ii) Now, let's move on to the second part. We want to find the probability that more than 73 out of 125 vehicles are cars.

The problem becomes a bit trickier here because we're dealing with a large sample size. However, we can use a normal approximation to estimate the probability.

First, we find the mean and standard deviation of the number of cars using the given percentages. Then, we can use these values to calculate the probability of selecting more than 73 cars using the normal distribution.

However, I must warn you that my calculations might be as accurate as an elephant on roller skates. So, take these approximations with a pinch of humor!

## To find the probability that fewer than 3 vehicles are buses in a random sample of 11 vehicles, we need to calculate the probability of selecting 0, 1, or 2 buses.

(i) The probability of selecting a bus in the random sample is 16% or 0.16. This means the probability of selecting a non-bus vehicle is 1 - 0.16 = 0.84.

To find the probability of selecting exactly 0 buses, we use the binomial probability formula: P(X = k) = (n C k) * p^k * (1 - p)^(n - k)

P(X = 0) = (11 C 0) * 0.16^0 * 0.84^(11 - 0)
= 1 * 1 * 0.84^11
â‰ˆ 0.1345

To find the probability of selecting exactly 1 bus, we use the same formula:

P(X = 1) = (11 C 1) * 0.16^1 * 0.84^(11 - 1)
= 11 * 0.16 * 0.84^10
â‰ˆ 0.3066

To find the probability of selecting exactly 2 buses:

P(X = 2) = (11 C 2) * 0.16^2 * 0.84^(11 - 2)
= 55 * 0.16^2 * 0.84^9
â‰ˆ 0.3432

Finally, we add up these probabilities to get the probability that fewer than 3 vehicles are buses:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
â‰ˆ 0.1345 + 0.3066 + 0.3432
â‰ˆ 0.7843

Therefore, the probability that fewer than 3 vehicles are buses in a random sample of 11 vehicles is approximately 0.7843.

(ii) To find the probability that more than 73 vehicles are cars in a random sample of 125 vehicles, we can use a normal approximation.

The mean number of cars in a sample of 125 vehicles can be calculated as follows:

mean = total number of vehicles * proportion of cars
= 125 * (1 - 0.20 - 0.16)
= 125 * 0.64
= 80

The standard deviation of the number of cars in a sample of 125 vehicles can be calculated using the formula:

standard deviation = sqrt(total number of vehicles * proportion of cars * proportion of non-cars)
= sqrt(125 * 0.64 * 0.20)
= sqrt(16)
= 4

We want to find the probability of selecting more than 73 cars, which is equivalent to finding the probability of selecting 74, 75, ..., or 125 cars.

Using the normal approximation, we can calculate the probability using the standard normal distribution:

P(X > 73) = P(Z > (73 - 80) / 4)
= P(Z > -1.75)

Using the standard normal table or a calculator, we find that P(Z > -1.75) is approximately 0.9599.

Therefore, the probability that more than 73 vehicles are cars in a random sample of 125 vehicles, using a suitable approximation, is approximately 0.9599.

## To solve these probability problems, we first need to understand how to calculate probabilities in this scenario.

Let's break down the problem into two parts:

(i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses.

To solve this, we first need to find the probability of selecting 0, 1, or 2 buses from the sample.

1. Calculate the probability of selecting 0 buses:
P(0 buses) = (100% - 16% cars) * (100% - 20% trucks)
= 84% cars * 80% trucks
= 0.84 * 0.80

2. Calculate the probability of selecting 1 bus:
P(1 bus) = 16% cars * (100% - 16% cars) * (100% - 20% trucks)
= 0.16 * 0.84 * 0.80

3. Calculate the probability of selecting 2 buses:
P(2 buses) = 16% cars * 16% cars * (100% - 20% trucks)
= 0.16 * 0.16 * 0.80

Finally, calculate the sum of the three probabilities to get the probability of selecting fewer than 3 buses:
P(fewer than 3 buses) = P(0 buses) + P(1 bus) + P(2 buses)

(ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the probability that more than 73 are cars.

To solve this, we can use the normal distribution as an approximation. Since the sample size is large (125), the sample proportion (cars) will follow approximately a normal distribution.

1. Calculate the mean of the sample proportion:
Mean = sample size * proportion of cars
= 125 * (100% - 16% trucks - 20% buses)
= 125 * (100% - 16% - 20%)
= 125 * 64%

2. Calculate the standard deviation of the sample proportion:
Standard Deviation = sqrt(sample size * proportion of cars * (100% - proportion of cars))
= sqrt(125 * 64% * 36%)

Now, we want to find the probability of selecting more than 73 cars:

1. Calculate z-score:
z = (73 - Mean) / Standard Deviation

2. Use a standard normal distribution table or calculator to find the probability associated with the calculated z-score.

Note: For approximation using the normal distribution, we assume the sample is random, independent, and the population is large enough.

By following the calculations described above, you should be able to find the probabilities for both parts (i) and (ii) of the question.