probability

#1
a pharmaceutical firm claims that a certain medication is 92% effective in relieving arthric pain. if ten people who suffer from arthritis are randomly selected and given the medicine, and assuming that the claim is correct, find the probability that

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
 
#2
I wish we had LaTex. It would be handy for this problem.

10
#1. sum(C(10,k)p^kq^(10-k)), p=0.92 and q=0.08
k=8


#2. C(10,2)p^2q^8
 
#4
pls help i dont think im on the right track

A)
sum(C(10,8) 0.92 ^8 X 0.08^(10-8)), p=0.92 and q=0.08 k=8
7.36 X 0.64(2)=9.42

b) C(10,2)0.92^2X0.92 ^8=

I THINK IM REALLY REALLY LOST.
 

JohnM

TS Contributor
#5
(a)
use formula for r = 8, 9, 10 and sum them:
C(10,r) * 0.92^r * 0.08^(10-r)



(b)
C(10,2) * 0.08^2 * 0.92^8
 
#7
Checking to see if im correct

Hi, when i see something like this (10,2) does that mean i should multiply. i dont know what the comma means.

For question A i got 0.410
 
#8
Checking If Im Correct

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
 
#10
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
NO, you're not correct. You're looking for probability. It must be between 0 and 1. Think about it, 3768.32?,

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
 

JohnM

TS Contributor
#13
Galactus,
Take it easy. We're trying to help out beginners here.


=======================================

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
 
#16
Ok, i worked out the problem and i now understand. Thank you very much. I was having a really hard time with the first part and i think thats what was messing me up.

THANK AGAIN. :)
 
#17
Ok, for question 2.
i tried it and i multiply 0.0064*0.51321*2=6.569
I hope this is correct. Im sure im not suppose to get 6.

The second thing i was going to do was to subtract 2 from which then gives me 6. i

Am i on the right path.