Question 663913: You pick 5 apples from a grocery store owned by the professor. The probability of a randomly chosen apple being rotten is 0.518. Let X represent the number of apples that are NOT rotten. Find the probability of X=k in the table below.
k P(X = k)
0
1
2
3
4
5
I tried multiplying .482*0 then .482*1 all the way down the table.
Found 3 solutions by solver91311, ewatrrr, jim_thompson5910:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The probability of  successes in  trials where  is the probability of success on any given trial is given by:
\ =\ {{n}\choose{k}}\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where  is the number of combinations of things taken at a time and is calculated by !}) . Hint:  . Also 
You want:
\ =\ {{5}\choose{0}}\left(0.518\right)^0\left(1\,-\,0.518\right)^{5\,-\,0})
Then
\ =\ {{5}\choose{1}}\left(0.518\right)^1\left(1\,-\,0.518\right)^{5\,-\,1})
and so on.
Happy calculator button punching. If you have Excel on the PC or Numbers on the Mac, then enter =BINOMDIST(k, n, p, FALSE) replacing k, n, and p with the appropriate values. The false means "not cumulative" and causes the function to return the probability of exactly k successes. If you use TRUE, you get the probability of k or less. =1 - BINOMDIST(k - 1, n, p, TRUE) gives you the probability of at least k successes.
John
My calculator said it, I believe it, that settles it
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi,
Note: The probability of x successes in n trials is:
P = nCx*  where p and q are the probabilities of success and failure respectively.
In this case p = .482 & q(rotton) = .518 and n = 5 nCx = 
Let X represent the number of apples that are NOT rotten.
P(X = 0) = 5C0(.482)^0(.518)^5
P(X = 1) = 5C1(.482)^1(.518)^4
P(X = 2) = 5C2(.482)^2(.518)^3
P(X = 3) = 5C3(.482)^3(.518)^2
P(X = 4) = 5C4(.482)^4(.518)^1
P(X = 5) = 5C5(.482)^5(.518)^0
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website! The probability of a randomly chosen apple being rotten is 0.518
So the probability of a randomly chosen apple NOT being rotten is 1 - 0.518 = 0.482
So p = 0.482
There are 5 apples picked, so n = 5.
This is a binomial distribution problem, so use the formula P(X = k) = (n C x)*(p)^(x)*(1-p)^(n-x)
For k = 0
P(X = k) = (n C x)*(p)^(x)*(1-p)^(n-x)
P(X = 0) = (5 C 0)*(0.482)^(0)*(1-0.482)^(5-0)
P(X = 0) = (5 C 0)*(0.482)^(0)*(0.518)^(5-0)
P(X = 0) = (1)*(0.482)^(0)*(0.518)^5
P(X = 0) = (1)*(1)*(0.037294844329568)
P(X = 0) = 0.037294844329568
P(X = 0) = 0.03729
For k = 1
P(X = k) = (n C x)*(p)^(x)*(1-p)^(n-x)
P(X = 1) = (5 C 1)*(0.482)^(1)*(1-0.482)^(5-1)
P(X = 1) = (5 C 1)*(0.482)^(1)*(0.518)^(5-1)
P(X = 1) = (5)*(0.482)^(1)*(0.518)^4
P(X = 1) = (5)*(0.482)*(0.071997768976)
P(X = 1) = 0.17351462323216
P(X = 1) = 0.17351
For k = 2
P(X = k) = (n C x)*(p)^(x)*(1-p)^(n-x)
P(X = 2) = (5 C 2)*(0.482)^(2)*(1-0.482)^(5-2)
P(X = 2) = (5 C 2)*(0.482)^(2)*(0.518)^(5-2)
P(X = 2) = (10)*(0.482)^(2)*(0.518)^3
P(X = 2) = (10)*(0.232324)*(0.138991832)
P(X = 2) = 0.32291138377568
P(X = 2) = 0.32291
For k = 3
P(X = k) = (n C x)*(p)^(x)*(1-p)^(n-x)
P(X = 3) = (5 C 3)*(0.482)^(3)*(1-0.482)^(5-3)
P(X = 3) = (5 C 3)*(0.482)^(3)*(0.518)^(5-3)
P(X = 3) = (10)*(0.482)^(3)*(0.518)^2
P(X = 3) = (10)*(0.111980168)*(0.268324)
P(X = 3) = 0.30046966598432
P(X = 3) = 0.30047
For k = 4
P(X = k) = (n C x)*(p)^(x)*(1-p)^(n-x)
P(X = 4) = (5 C 4)*(0.482)^(4)*(1-0.482)^(5-4)
P(X = 4) = (5 C 4)*(0.482)^(4)*(0.518)^(5-4)
P(X = 4) = (5)*(0.482)^(4)*(0.518)^1
P(X = 4) = (5)*(0.053974440976)*(0.518)
P(X = 4) = 0.13979380212784
P(X = 4) = 0.13979
and finally, for k = 5
P(X = k) = (n C x)*(p)^(x)*(1-p)^(n-x)
P(X = 5) = (5 C 5)*(0.482)^(5)*(1-0.482)^(5-5)
P(X = 5) = (5 C 5)*(0.482)^(5)*(0.518)^(5-5)
P(X = 5) = (1)*(0.482)^(5)*(0.518)^0
P(X = 5) = (1)*(0.026015680550432)*(1)
P(X = 5) = 0.026015680550432
P(X = 5) = 0.02602
The updated table is then
| k |
P(X = k) |
| 0 |
0.03729 |
| 1 |
0.17351 |
| 2 |
0.32291 |
| 3 |
0.30047 |
| 4 |
0.13979 |
| 5 |
0.02602 |
How to read the table above. Let's say you look at the third row. In this row is 2 and 0.32291, which says "the probability of picking exactly 2 rotten apples (out of 5 total) is roughly 0.32291 or 32.291%
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