Approximately 13% of the population are left-handed. If three people are randomly selected, what is the...

Approximately 13% of the population are left-handed.

If three people are randomly selected, what is the probability that all are left- handed? (round to 4 decimal places)

If three people are randomly selected, what is the probability that at least 1 person is not left handed? (round to 4 decimal places)

Notice that these two events are complementary (the probabilities do add up to 1).

Expert Solution

n = 3, p = 0.13

Since it is a binomial distribution, so P(X = x) = nCx * p^x * (1 - p)^(n - x)

a)P(all are left handed) = P(X = 3) = 3C3 * (0.13)^3 * (0.87)^0 = 0.0022

b) P(at least one person is not left handed) = 1 - P(all are left handed)

= 1 - 0.0022

= 0.9978

orchestra answered 2 years ago

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