Question

Police estimate that​ 25% of drivers drive without their seat belts. If they stop 6 drivers at​ random, find the probability that all of them are wearing their seat belts. Round your answer to four decimal places.

Answer 1

Answer: The required probability is 0.178.

Step-by-step explanation:

Since we have given that

Probability of drivers drive without their seat belts = 25%

Probability of drivers drive with their seat belts = 100-25=75%

Sample = Number of drivers = 6

Using "Binomial distribution", we get that

So, Probability that all of them are wearing their seat belts is given by

$P(X=6)=^6C_6(0.75)^6(0.25)^0=(0.75)^6=0.178$

Hence, the required probability is 0.178.

Answer 2

Answer:

17.80% probability that all of them are wearing their seat belts.

Step-by-step explanation:

For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal 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.

Police estimate that​ 25% of drivers drive without their seat belts.

This means that 75% wear their seatbelts, so $p = 0.75$

If they stop 6 drivers at​ random, find the probability that all of them are wearing their seat belts.

This is $P(X = 6)$.

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

$P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.1780$

17.80% probability that all of them are wearing their seat belts.