Thread: Binomial Probability table

Cars make up 70% of a country's registered vehicles. Consider a random sample of 20 vehicles.
What is the probability that all are cars?

The binomial probability table I'm using only goes up to p = 0.5 and for this question p = 0.7%

Here's what I've come up with:
P(X = x), so x is 20 = P(Y = 20 - 20), so Y is 0. Y ~ B(20, 0.3) - I did 1 minus 0.7. So I find that on the table and the answer is 0. So is that the answer?

Originally Posted by brumby_3

Cars make up 70% of a country's registered vehicles. Consider a random sample of 20 vehicles.
What is the probability that all are cars?

The binomial probability table I'm using only goes up to p = 0.5 and for this question p = 0.7%

Here's what I've come up with:
P(X = x), so x is 20 = P(Y = 20 - 20), so Y is 0. Y ~ B(20, 0.3) - I did 1 minus 0.7. So I find that on the table and the answer is 0. So is that the answer?

Let X be the random variable number of cars.

X ~ Binomial(n = 20, p = 0.7)

Pr(X = 20) = (0.7)^20 = 0.0008 (correct to 4 decimal places).