Consider a multiple-choice question exam consisting of 20 questions. Assume that each question has four possible choices, and only one of them is correct. If a student is going to guess solutions at the exam, what is the probability that he answers at most three of them are correct

0.22514

Step-by-step explanation:

Given :

1 answer from 4 options

p = 1/4 = 0.25 ; q = 1 - 0.25 = 0.75

n = 20

P(x ≤ 3) = p(x =0) +p(x=1) + p(x=2) +p(x=3)

P(x = x) = nCr * p^x * q^(n-r)

p(x =0) = 20C0 * 0.25^0 * 0.75^20 = 0.00317

p(x =1) = 20C1 * 0.25^1 * 0.75^19 = 0.02114

p(x =2) = 20C2 * 0.25^2 * 0.75^18 = 0.06694

p(x =3) = 20C3 * 0.25^3 * 0.75^17 = 0.13389

P(x ≤ 3) = 0.00317+0.02114+0.06694+0.13389 = 0.22514

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )

51-50

step-by-step explanation: