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a poll reported that 80% of adults were satisfied with the job major airlines were doing. suppose 12 adults are selected at random and the number who are satisfied are recorded. find the probability that a)9 or more are satisfied with the airlines b)between 3 and 6 adults, inclusive, were satisfied

There are two parts to the question.

Only two possible scenarios: a) success or b) failure

Applying the binomial concept:

P(satisfaction) = 80% or 0.8; P(dissatisfaction) = 20% or 0.2

Therefore, P(9 or more satisfied persons) = P(9) + P(10) + P(11) + P(12)

1st part

= 12C9 * (0.8)^9 * (0.2)^3 + 12C10 * (0.8)^10 * (0.2)^2 + 12C11 * (0.8)^11 * (0.2)^1 + 12C12 * (0.8)^12 * (0.2)^ 0

= 0.2362 + 0.2835 + 0.2062 + 0.0687

= 0.7946 (1st part of the puzzle)

2nd part.

Solve for p(3, 4, 5, and 6 satisfied persons)

12C3*(0.8)^3 * (0.2)^9 + 12C4*(0.8)^4 * (0.2)^8 + 12C5*(0.8)^5 * (0.2)^7 + 12C6* (0.8)^6* (0.2)^6

= 0.00005767 + 0.00051904 + 0.0033219 + 0.0154995

= 0.01940

You can use binomial probability because there are only two outcomes.

a) 12C9 (.8)^9 (.2)^3 + 12C10 (.8)^10 (.2)^2 + 12C11 (.8)^11 (.2) + (.8)^12 = .79457

b) 12C3 (.8)^3 (.2)^9 + 12C4 (.8)^4 (.2)^8 + 12C5 (.8)^5 (.2)^7 + 12C6 (.8)^6 (.2)^6 = .01940