is at least one corect answer.A student makes random guesses for the the ten true-false question on a quiz.Find the probability that there i
Another way of wording this is that they're not all wrong. So you're looking for 1 - P(all wrong), which is 1 - (1/2)^10.A student makes random guesses for the the ten true-false question on a quiz.Find the probability that there i
Let X be the number of correct answers. X has the binomial distribution with n = 10 trials and success probability p = 0.5
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.
The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.
X ~ Binomial( n = 10 , p = 0.5 )
the mean of the binomial distribution is n * p = 5
the variance of the binomial distribution is n * p * (1 - p) = 2.5
the standard deviation is the square root of the variance = 鈭?( n * p * (1 - p)) = 1.581139
The Probability Mass Function, PMF,
f(X) = P(X = x) is:
P( X = 0 ) = 0.0009765625
P( X = 1 ) = 0.009765625
P( X = 2 ) = 0.04394531
P( X = 3 ) = 0.1171875
P( X = 4 ) = 0.2050781
P( X = 5 ) = 0.2460938
P( X = 6 ) = 0.2050781
P( X = 7 ) = 0.1171875
P( X = 8 ) = 0.04394531
P( X = 9 ) = 0.009765625
P( X = 10 ) = 0.0009765625
The Cumulative Distribution Function, CDF,
F(X) = P(X 鈮?x) is:
x
鈭?P(X = t) =
t = 0
P( X 鈮?0 ) = 0.0009765625
P( X 鈮?1 ) = 0.01074219 = 1- P(X = 0) %26lt;%26lt;%26lt; ANSWER
P( X 鈮?2 ) = 0.0546875
P( X 鈮?3 ) = 0.171875
P( X 鈮?4 ) = 0.3769531
P( X 鈮?5 ) = 0.6230469
P( X 鈮?6 ) = 0.828125
P( X 鈮?7 ) = 0.9453125
P( X 鈮?8 ) = 0.9892578
P( X 鈮?9 ) = 0.9990234
P( X 鈮?10 ) = 1
P(at least 1 correct) = 1 - P(none are correct)
the probability that none of the 10 answers are correct is (1/2)^10
P = 1 - (1/2)^10
P 鈮?0.9990234375
edit:
1023/1024 鈮?0.9990234375
so the answer is A
so the only exception to at least one right is all wrong. The probability of this is (1/2)^10. You subtract this from 1 to since it is what we don't want, and the probability is 1023/1024
P = 1/2^10
= 0.0009765625
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