If a student was randomly guessing while completing a 20 question multiple choice exam, what is the probability they would score
a 50% or higher
1 answer:
Answer:
you want 4 correct and 16 incorrect
there are 20 questions
each question has four answers, so
P(right answer) = 1/4
P(wrong answer) = 3/4
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Since you want 4 correct of 20 we have a combination of 20C4
This is a binomial problem where p = 1/4, q = 3/4 and we get
(20 "choose" 4)*(probability correct)^(number correct)*(probability incorrect)^(number incorrect)
putting numbers in we get
(20c4)*(1/4)^4*(3/4)^16
This gives us
~ .189685
Step-by-step explanation:
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