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August 20, 2014

August 20, 2014

Posted by **jimmy** on Sunday, July 22, 2012 at 5:11pm.

a) if a random sample of 5 consumer complaints is selected, what is the probability that all 5 were resolved?

b)if a random sample of 5 consumer complaints is selected, what is the probability that at least 1 was not resolved?

c) if a random sample of 5 consumer complaints is selected, what is the probability that none were resolved?

d) if a random sample of 5 consumer complaints is selected, what is the probability that at least 1 was resolved?

- statistics -
**MathMate**, Sunday, July 22, 2012 at 5:30pmUse binomial theorem with N=5, p=0.70

=> q=1-p=0.3

Let

C(n,r)=binomial coefficient=n!/(r!(n-r)!)

(a) all five resolved

P(X=5)=C(5,5)p^5q^0=1*0.7^5*1=0.17

(b) at least one not resolved

i.e. not all five

P(X≤4)=1-P(X=5)

(c) none resolved

P(X=0)=C(5,0)p^0q^5=1*1*0.3^5=0.002

(d) at least one resolved

i.e. not none resolved

P(X≥1)=1-P(X=0)

- statistics -
**MathMate**, Sunday, July 22, 2012 at 6:26pmA little suggest for follow-ups to my responses.

Start with a follow-up to the original post. If your patience runs out,*then*start a new post with or without the recipient's name under "school subject".

Assuming you have no problem with parts (a) and (c), parts (b) and (d) are simply the complement to the preceding part.

For example,

P(X≤4) means P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4).

However, we make use of the fact that the sum of probabilities of all possible events add up to 1, or

P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=1 (all possible events), so by transposition, we obtain:

P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

=1-P(X=5)

as was done in (b) and (d).

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