Question 475934
This is a binomial distribution problem. Recall that the pdf of a binomial distribution for n number of trials with probability p is


P(X = x) = (n C x)*(p)^x*(1-p)^(n-x)



For more help with binomial distributions, see <a href="http://stattrek.com/tables/binomial.aspx">this calculator</a>.



In this case, n = 10 (ie there are 10 trials since there are 10 questions) and p=1/2=0.5


a)


In part a), x = 10 (since we want to get all 10 correct)



P(X = 10) = (10 C 10)*(0.5)^(10)*(1-0.5)^(10-10) 



P(X = 10) = (10 C 10)*(0.5)^(10)*(0.5)^(10-10) 



Note: 10 C 10 = (10!)/(10!(10-10)!) = 1



P(X = 10) = (1)*(0.5)^(10)*(0.5)^0 



P(X = 10) = (1)*(0.0009765625)*(1) 



P(X = 10) = 0.0009765625



So the probability of getting all 10 correct is 0.0009765625 (which is roughly 0.098 % .... a very very small chance)

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b)


In part b), x = 7 (since we want exactly 7)



P(X = 7) = (10 C 7)*(0.5)^(7)*(1-0.5)^(10-7) 



P(X = 7) = (10 C 7)*(0.5)^(7)*(0.5)^(10-7) 



Note: 10 C 7 = (10!)/(7!(10-7)!) = 120



P(X = 7) = (120)*(0.5)^(7)*(0.5)^3 



P(X = 7) = (120)*(0.0078125)*(0.125) 



P(X = 7) = 0.1171875



So the probability of getting exactly 7 correct is 0.1171875 (which is roughly 11.72%)



For more help with binomial distributions, see <a href="http://stattrek.com/tables/binomial.aspx">this calculator</a>.