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uysha [10]
3 years ago
13

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

cles on the roads are occupied by just the driver. (Round your answers to four decimal places.) (a) If you choose 10 vehicles at random, what is the probability that more than half (that is, 6 or more) carry just one person?
Mathematics
1 answer:
Leviafan [203]3 years ago
7 0

Answer:

72.92% probability that more than half (that is, 6 or more) carry just one person.

Step-by-step explanation:

For each vehicle, there are only two possible outcomes. Either they carry only one person, or they carry more than one person. The vehicles are chosen at random, which means that the probability of a vehicle carrying just one driver is independent from other vehicles. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

64% of vehicles on the roads are occupied by just the driver.

This means that p = 0.64

(a) If you choose 10 vehicles at random, what is the probability that more than half (that is, 6 or more) carry just one person?

This is P(X \geq 6) when n = 10

We have that

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10).

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{10,6}.(0.64)^{6}.(0.36)^{4} = 0.2424

P(X = 7) = C_{10,7}.(0.64)^{7}.(0.36)^{3} = 0.2462

P(X = 8) = C_{10,8}.(0.64)^{8}.(0.36)^{2} = 0.1642

P(X = 9) = C_{10,9}.(0.64)^{9}.(0.36)^{1} = 0.0649

P(X = 10) = C_{10,10}.(0.64)^{10}.(0.36)^{0} = 0.0115

So

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2424 + 0.2462 + 0.1642 + 0.0649 + 0.0115 = 0.7292.

72.92% probability that more than half (that is, 6 or more) carry just one person.

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