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

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

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:

answer: 5100 just subtract

step-by-step explanation:

step-by-step explanation:

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