SOLUTION: use the binomial expression (p+q)^n to calculate a binomial distribution with n=5 and p=0.3. Can someone help me with this? I have to do it step by step and I'm not sure how to.

Question 1105264: use the binomial expression (p+q)^n to calculate a binomial distribution with n=5 and p=0.3.
Can someone help me with this? I have to do it step by step and I'm not sure how to. I will particate if you guide me.
Thanks in advance!
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
by inspecting Pascal's triangle, the following formula is apparent
:
(a+b)^n = a^n + (nC1)a^(n-1)b + (nC2)a^(n-1)q^2 +...+(nCn-1)ab^(n-1) + b^n
:
our n = 5 and p=0.3
:
for a binomial distribution q = 1-p, for this problem q = 0.7 and a = p and b = q
:
nCr = n! / (r! * (n-1)!)
:
you are asked to calculate each term for (0.3 + 0.7)^5
:
Note that each term is a probability of k failures in n trials
:
Probability (P) ( 0 failures in 5 trials ) = (.3)^5
Probability (P) ( 1 failure in 5 trials) = 5C1 * (.3)^4 * (.7)
Probability (P) ( 2 failures in 5 trials) = 5C2 * (.3)^3 * (.7)^2
Probability (P) ( 3 failures in 5 trials) = 5C3 * (.3)^2 * (.7)^3
Probability (P) ( 4 failures in 5 trials) = 5C3 * (.3) * (.7)^4
Probability (P) ( 5 failures in 5 trials) = (.7)^5
:

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