Question

Suppose that 3% 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?

Answer 1

Answer:   0.336

Step-by-step explanation:

Let x be the number of students who take the SAT each year receive special accommodations.

Probability that a student who take the SAT each year receive special accommodations =0.03

Here , $X\sim\text{ Bin(n=20, p=0.03)}$

Binomial probability formula : $P(X=x)^nC_xp^x(1-p)^{n-x}$ , where p =probability of each successes , n is total trials.

Then ,

$P(X=1)=^{20}C_1(0.03)^1(1-0.03)^{20-1}\\\\=(20)(0.03)(0.97)^{19}\approx0.336$

Hence, the  probability that exactly 1 received a special accommodation is 0.336 .