Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities. Consider a random sample of 20 students who have recently taken the test. (Round your probabilities to three decimal places.)a) What is the probability that exactly 1 received a special accommodation?
(b) What is the probability that at least 1 received a special accommodation?
(c) What is the probability that at least 2 received a special accommodation?

a.) 0.368

b.) 0.442

c.) 0.190

Step-by-step explanation:

This is solved by the probability distribution formula. p(X=r)

P(X=r) = nCr × p^r × q^n-r

Where,

n = total number of sampled outcomes

r = number of successful outcome among the sample

p = probability of success

q = probability of failure.

If n=20, p = 0.04, q = 0.96

a.) Probability that at exactly one special student received accommodation = P(X=1)

P(X=1) =20C1 × 0.04¹ × 0.96^19

P(X=1) = 0.368 (3dp)

b.) Probability that at least one special student received accommodation = 1 - [probability that no special student received accommodation.]

P(X=0) = 20C0 × 0.04^0 × 0.96^20

P(X=0) = 0.442 (3d.p)

Probability of at least 1 received accommodation = 1 - P(X=0)

= 1 - 0.442

= 0.558.

c.) Probability that at least two special students received accommodation = 1 - [P(X=1)] - [P(X=0)]

As determined earlier,

P(X=1) = 0.368

P(X=0) = 0.442 (3dp)

Hence, Probability that at least two special students get accommodation = 1 - 0.368 - 0.442 = 0.190 (3dp)

18a

step-by-step explanation:

simply because it is. it's the highest common factor