The probability that a vehicle will change lanes while making a turn is 45%. Suppose a random sample of 7 vehicles are observed making turns at a busy intersection. Find the probability that a least 5 vehicles will change lanes while making the turn.

Using the binomial distribution, it is found that there is a 0.1529 = 15.29% probability that a least 5 vehicles will change lanes while making the turn.

For each vehicle, there are only two possible outcomes, either they change lanes while making a turn, or they do not. The probability of a vehicle changing lanes while making a turn is independent of any other vehicle, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

The parameters are:

In this problem:

The probability that at least 5 vehicles will change lanes while making the turn is:

In which

Then

0.1529 = 15.29% probability that a least 5 vehicles will change lanes while making the turn.

To learn more about the binomial distribution, you can take a look at link

0.1529

Step-by-step explanation:

Given that:

Probability of changing lane ; p = 45% = 0.45

Sample size (number of trials) = n = 7

Probability that atleast 5 vehicles will change lane :

Using the binomial probability formula :

P(x = x) = nCx * p^x * (1 - p)^(n-x)

(1 - p) = 1 - 0.45 = 0.55

P(x ≥ 5) = p(x = 5) + p(x = 6) + p(x = 7)

P(x = 5) = 7C5 * 0.45^5 * 0.55^2

P(x = 5) = 21 * 0.45^5 * 0.55^2 = 0.117221

P(x = 6) = 7C6 * 0.45^6 * 0.55^1

P(x = 6) = 7 * 0.45^6 * 0.55^1 = 0.031969

P(x = 7) = 7C7 * 0.45^7 * 0.55^0

P(x = 7) = 1 * 0.45^7 * 0.55^0 = 0.003736

P(x ≥ 5) = p(x = 5) + p(x = 6) + p(x = 7)

P(x ≥ 5) = 0.117221 + 0.031969 + 0.003736

P(x ≥ 5) = 0.152926

P(x ≥ 5) = 0.1529

560

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