Matthew A. answered • 05/01/20
Ivy League STEM Tutoring and Test Prep Skills (MCAT 100th percentile)
This questions uses a binomial, distribution to calculate the probability of something happening x times our of a possible total of n.
a) The probability of at least 3 right is the probability of getting 3, 4, or 5 answers right. Notice the "or" we must sum these probabilities.
Pr(right = 3) = (5!)/[(3!)x(2!) x (.6)3(1-.6)2 = 0.3456
Pr(right = 4) = (5!)/[(4!)x(1!) x (.6)4(1-.6)1 = 0.2592
Pr(right = 5) = (5!)/[(5!)x(0!) x (.6)5(1-.6)0 = 0.7776
0.3456 + .07776 + .2592 = 0.68256
b) The probability of getting no more than 2 answers right, is the probability of getting, 0, 1 or 2 answer right. Notice the or once again, we must due the same process as above and calculate the individual probabilities of those outcomes and sum them.
However, A quicker method is to notice that the probability of getting 2 or less, is the same as NOT getting 3 or more, which we've calculated above. Let X be the number of questions we get right
Pr(X≤ 2) = 1 - Pr(X ≥ 3)
Pr(X≤ 2) = 1 - 0.68256
Pr(X≤ 2) = 0.31744
Hope that helps :)
