SOLUTION: It is found that 60% of American victims of healthcare fraud are senior citizens. If 10 victims are randomly selected, what is the probability that at least 6 are senior citizens

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Question 1146838: It is found that 60% of American victims of healthcare fraud are senior citizens.
If 10 victims are randomly selected, what is the probability that at least 6 are senior citizens?

Answer by ikleyn(52835) About Me  (Show Source):
You can put this solution on YOUR website!
.
This is a binomial distribution type problem where the probability under the question is the sum


     P = sum+%28C%2810%2Ck%29%2Ap%5Ek%2Aq%5E%2810-k%29%2Ck=6%2C10%29      (1)


The number of trials is              10;
The indexes of success trials        k = 6,7,8,9,10
The probability of success trial     p = 0.6;
                                     q = 1 - p
C(n,k) = n! / (k! * (n-k)!)          are binomial coefficients.


The sum  (1)  is equal to  1 - sum%28C%2810%2Ck%29%2Ap%5Ek%2Aq%5E%2810-k%29%2Ck=0%2C5%29.     (2)



Instead of calculating every term of (2) manually and then summing them up, you may use Excel function 

BINOM.DIST(5, 10, 0.6, TRUE)  to calculate the value


    sum%28C%2810%2Ck%29%2Ap%5Ek%2Aq%5E%2810-k%29%2Ck=0%2C5%29 = 0.366897.    


Therefore, the value of  (2)  is equal to  1 - 0.366897 = 0.633103 (approximately).    ANSWER

On Excel function BINOM.DIST, see its description everywhere, for example

https://support.office.com/en-us/article/binom-dist-function-c5ae37b6-f39c-4be2-94c2-509a1480770c