Using the concept of binomial probability, the probability that 3 or more of the drivers are uninsured is 0.062
Recall :
P(x = x) = nCx * p^x * q^(n-x)
Where ;
n = number of trials = 7
x ≥ 3
p = probability of success = P(uninsured) = 0.14
q = 1 - p = 1 - 0.14 = 0.86
P(x ≥ 3) = P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7)
Using a binomial probability calculator :
P(x = 3) = 0.05253
P(x = 4) = 0.0086
P(x = 5) = 0.0008
P(x = 6) = 0.000045
P(x = 7) = 0.000001
Hence,
P(x ≥ 3) = 0.05253 + 0.0086 + 0.0008 + 0.000045 + 0.000001 = 0.062 (3 decimal places)
P(x ≥ 3) = 0.062 (3 decimal places).
Therefore, the probability that ≥ 3 of the drivers are uninsured is 0.062.
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