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)