Okay, so I'm studying for my math test and I have NO CLUE how to do some of these problems.
The situation is that a girl is taking a test with 5 questions. Each question has 5 answers listed "a-e". A new student is taking a test on his first day of school and has no prior knowledge of the material. (So he is COMPLETELY guessing)
What is the probability that he only misses 2 questions?
What is the probability that he misses more than 2 questions?
M3
Favorite Answer
to get this problem, you must be knowing about the binomial distribution.
the one here has parameters n = 5, p = 1/5, q = 1- p = 4/5
formula for k successes in n trials is
P[k] = nCk * p^k * q^(n-k)
qa
miss 2 questions means get 3 right,
P[3] = 5C3 * [1/5]^3 * [4/5]^2
= 10*1*16/3125
= 32/625 or 5.12%
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qb
misses more than 2 q's means gets 0,1 or 2 right
calculate P[0] + P[1] + P[2]
= 5C0*[1/5]^0 *[4/5]^5 + 5C1*[1/5][4/5]^4 + 5C2*[1/5]^2 *[4/5]^3
= [4^5 + 5*4^4 + 10*4^3] / 5^5
= 2944 / 3125 or 94.208%
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