The probability of getting 3 or more who were involved in a car accident last year is 0.126.
Given 9% of the drivers were involved in a car accident last year.
We have to find the probability of getting 3 or more who were involved in a car accident last year if 14 were selected randomly.
We have to use binomial theorem which is as under:
n
where p is the probability an r is the number of trials.
Probability that 3 or more involved in a car accident last year if 14 are randomly selected=1-[P(X=0)+P(X=1)+P(X=2)]
=1-{
}
=1-{1*0.2670+14!/13!*0.9*0.29+14!/2!12!*0.0081*0.2358}
=1-{0.2670+0.3654+0.2358}
=1-0.8682
=0.1318
Among the options given the nearest is 0.126.
Hence the probability that 3 or more are involved in the accident is 0.126.
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28/42 is 66.66…%
0.66…(24) = 16.
Answer: sixteen red velvets
It depends if it is negative that's being timesed by 2 more negatives because then that would make it negative but if it's positive times a negatives times a negative then it would be positive and no matter what a positive times any positives and no negatives that will be positive. Hope this helped! :)
Answer:
What does this mean lolzers
Step-by-step explanation:
True
Step-by-step explanation:
explanation is written inside the attachment