A) At least eight of them will be relieved of their arthritic pain

B) Exactly two of them will not be relieved of their arthritic pain

- Thread starter sweetlady
- Start date

A) At least eight of them will be relieved of their arthritic pain

B) Exactly two of them will not be relieved of their arthritic pain

http://www.talkstats.com/showthread.php?t=146

sweetlady said:

Can someone pls tell me if im correct with this answer. I worked it out again and for

A) I GOT

80*7.36*6.4=3768.32

PLS JUST TELL ME IF IM CORRECT OR NOT

A) I GOT

80*7.36*6.4=3768.32

PLS JUST TELL ME IF IM CORRECT OR NOT

C(n,r) means combinations. A selection of r objects from a group of n objects without regard to order. These are the coefficients in your binomial expansion. Given by n!/[(n-r)!r!].

Since they ask for at least 8, they want 8,9, and 10.

You have: C(10,8)p^8q^2+C(10,9)p^9q+C(10,10)p^10, where p=0.92 and q=0.08

Take it easy. We're trying to help out beginners here.

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Sweetlady,

Whenever you see C(n,r) it is short-hand for the formula for the number of combinations of n items, taking r at a time:

n!/[(n-r)!r!]

So, for C(10,8), we get: 10! / [ (10 - 8)! * 8! ] which computes to 45

for C(10,9), we get: 10! / [ (10 - 9)! * 9! ] which computes to 10

for C(10,10), we get: 10! / [ (10 - 10)! * 10! ] which computes to 1

So, to compute the probability of "at least 8" we add up the probability of 8, 9, and 10

P(r) = C(n,r) * p^r * q^(n-r)

P(8) = C(10,8) * 0.92^8 * 0.08^2 = .1478

P(9) = C(10,9) * 0.92^9 * 0.08^1 = .3778

P(10) = C(10,10) * 0.92^10 * 0.08^0 = .4344

So, P(at least 8) = .1478 + .3778 + .4344 = .96