Question

For Questions No. 2-4 use the following information. Based on a USA Today poll, 15% of the population believes that college is no longer a good investment. Among 10 randomly selected people, 2 believe that college is no longer a good investment. (2) Assuming 15% rate of belief, find the probability that, in 10 people, at least two believe that college is no longer a good investment (A) $P(x = 2) = \binom{10}{2} \times 0.15^2 \times (0.85)^8$ (B) $P(x < 2) = P(0) + P(1) = (0.85)^{10} + \binom{10}{1} \times (0.15) \times (0.85)^9$ (C) $P(x > 2) = 1 - P(x \le 2) = 1 - [(0.85)^{10} + \binom{10}{1} \times (0.15) \times (0.85)^9 + \binom{10}{2} \times (0.15)^2 \times (0.85)^8]$ (D) $P(x \ge 2) = 1 - P(x < 2) = 1 - [(0.85)^{10} + \binom{10}{1} \times (0.15) \times (0.85)^9]$

          For Questions No. 2-4 use the following information. Based on a USA Today poll, 15% of the population believes that college is no longer a good investment. Among 10 randomly selected people, 2 believe that college is no longer a good investment.
(2) Assuming 15% rate of belief, find the probability that, in 10 people, at least two believe that college is no longer a good investment
(A) $P(x = 2) = \binom{10}{2} \times 0.15^2 \times (0.85)^8$
(B) $P(x < 2) = P(0) + P(1) = (0.85)^{10} + \binom{10}{1} \times (0.15) \times (0.85)^9$
(C) $P(x > 2) = 1 - P(x \le 2) = 1 - [(0.85)^{10} + \binom{10}{1} \times (0.15) \times (0.85)^9 + \binom{10}{2} \times (0.15)^2 \times (0.85)^8]$
(D) $P(x \ge 2) = 1 - P(x < 2) = 1 - [(0.85)^{10} + \binom{10}{1} \times (0.15) \times (0.85)^9]$

For Questions No. 2-4 use the following information. Based on a USA Today poll, 15% of the population believes that college is no longer a good investment. Among 10 randomly selected people, 2 believe that college is no longer a good investment.
(2) Assuming 15% rate of belief, find the probability that, in 10 people, at least two believe that college is no longer a good investment
(A) P(x = 2) = 102× 0.15^2 × (0.85)^8
(B) P(x < 2) = P(0) + P(1) = (0.85)^10 + 101× (0.15) × (0.85)^9
(C) P(x > 2) = 1 - P(x ≤ 2) = 1 - [(0.85)^10 + 101× (0.15) × (0.85)^9 + 102× (0.15)^2 × (0.85)^8]
(D) P(x ≥ 2) = 1 - P(x < 2) = 1 - [(0.85)^10 + 101× (0.15) × (0.85)^9]

Added by Julie M.

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