Question 930996
This is a binomial distribution question.


n = 8
p = 1/2 = 0.5


Let's define the function B(k) to be the probability where the student misses k problems where {{{0<=k<=8}}}. So it's defined in terms of the <a href="http://stattrek.com/probability-distributions/binomial.aspx">binomial distribution formula</a>


B(k) = (n C k)*p^k*(1-p)^(n-k)


Let's plug in the constants n = 8 and p = 0.5 to get


B(k) = (8 C k)*0.5^k*(1-0.5)^(8-k)
B(k) = (8 C k)*0.5^k*0.5^(8-k)



Since it asks "what is the probability of missing no more than 1 question" we want to find P(X <= 1) which means we need to evaluate B(0) and B(1). Then we add up those two values


{{{P(X <= 1) = B(0) + B(1)}}}


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P(Missing 0) = B(0) = (8 C 0)*(0.5^0)*(0.5)^(8-0) = 0.00390625


P(Missing 1) = B(1) = (8 C 1)*(0.5^1)*(0.5)^(8-1) = 0.03125


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Now add up the two individual probabilities (PDF's) to get the cumulative probability (CDF)


B(0) + B(1) = 0.00390625+0.03125 = 0.03515625


P(X <= 1) = 0.03515625


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So the probability of missing no more than 1 question is <font color="red">0.03515625</font>



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