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There are 4c2 ways to get two correct, 4c3 ways to get 3 correct, and 4c4 ways to get all of them correct.
There are 2^4 total possible ways to answer the test.
The probability of getting at least two questions correct, then, is
(4c2 + 4c3 + 4c4) / (2^4)
= (6 + 4 + 1) / 16
= 11/16
There are 2^4 total possible ways to answer the test.
The probability of getting at least two questions correct, then, is
(4c2 + 4c3 + 4c4) / (2^4)
= (6 + 4 + 1) / 16
= 11/16
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Binomial distribution is used.
P(correct) = p = 1/2
q = 1-p = 1-1/2 = 1/2
n = 4
The general form of binomial distribution for deterrmining the probability of r successes is
P(r) = nCr*q^(n-r)*p^r
Required probability = P(r >/= 2) = P(r=2)+P(r=3)+P(r=4)
or 1 - P(r=0) - P(r=1)
= 1 - 4C0*(1/2)^4*(1/2)^0 - 4C1*(1/2)^3*(1/2)^1
= 1 - 1*1/16*1 - 4*1/8*1/2
= 1- 1/16 - 4/16
= 1 - 5/16
= 11/16
P(correct) = p = 1/2
q = 1-p = 1-1/2 = 1/2
n = 4
The general form of binomial distribution for deterrmining the probability of r successes is
P(r) = nCr*q^(n-r)*p^r
Required probability = P(r >/= 2) = P(r=2)+P(r=3)+P(r=4)
or 1 - P(r=0) - P(r=1)
= 1 - 4C0*(1/2)^4*(1/2)^0 - 4C1*(1/2)^3*(1/2)^1
= 1 - 1*1/16*1 - 4*1/8*1/2
= 1- 1/16 - 4/16
= 1 - 5/16
= 11/16
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Well if each question has two possible answers then that's 50% chance of getting it right. If theirs 4 questions then you have 200% chance of getting them all right and 100% chance of getting at least 2 correct.