Using it's concept, it is found that there is a 0.64 = 64% probability that neither flight is delayed.
<h3>What is a probability?</h3>
A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.
In this problem, there are 2 flights, each with a 0.8 probability of being on time. The flights are independent, hence the probability that both are on time is given by:
p = 0.8 x 0.8 = 0.64.
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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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Answer:
9
Step-by-step explanation:
18 ÷ {[(14) - (13)] + 1}
18 ÷ {1+1} =
18 ÷ 2 = 9
Hope that helps!
C. Because all the have to do is see what the multiply of both numbers are.