Question 730163
All 3 parts are dealing with the binomial distribution


In each case


n = 10
p = 0.4


a)


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


P(X = 5) = (10 C 5)*(0.4)^(5)*(1-0.4)^(10-5)


P(X = 5) = (10 C 5)*(0.4)^(5)*(0.6)^(10-5)


P(X = 5) = (252)*(0.4)^(5)*(0.6)^5


P(X = 5) = (252)*(0.01024)*(0.07776)


P(X = 5) = <font color="red">0.200658</font>


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b)


Use the same formula to find P(X = 4), P(X = 5), ..., all the way up to P(X = 8)


The steps will be the same as in part a (for each individual probability). However, the value of k will change. Doing that will give you this


P(X = 4)= 0.250823
P(X = 5)= 0.200658
P(X = 6)= 0.111477
P(X = 7)= 0.042467
P(X = 8)= 0.010617


Add them up to get...


P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) 


0.250823 + 0.200658 + 0.111477 + 0.042467 + 0.010617 


0.616042



So the answer to part b) is <font color="red">0.616042</font>




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c)


Same as part b), but now we want P(X = 8),..., up to P(X = 10).



P(X = 8)= 0.010617
P(X = 9)= 0.001573
P(X = 10)= 0.000105



P(X = 8) + P(X = 9) + P(X = 10)


0.010617 + 0.001573 + 0.000105


0.012295


So the answer to part c) is <font color="red">0.012295</font>