Question

Component A is broken by a probability of 0.05, while component B is broken by a probability of 0.10. Let X_a and X_b be the number of usable compnents of each.

What is the probability of acquiring more working (that is, not broken) A compnents than B components if the number of total components for A and B is n_A = 1200, n_B = 1300.

I guess this problem is related to binomal probability.

Answer 1

Is this a problem from a book? I am having trouble understanding the english.

Answer 2

Direct translation from the problem;

A product is composed of two components, A and B. The component A has a 0.05 chance of being faulty, while the corresponding probability for component B is 0.10. Assume the components are faulty independent on each other. You manufacture a total of $n_A$ of the component A and $n_B$ of the component B.

Calculate the probability that you get more usable A-components than B-components if $n_A$ = 1200 and $n_B$=1300

Answer 3

One way to do this, since we want more A working components than B working components, you can list the possibilities, starting with A = 1, (since A has to be greater than B, A cannot equal to zero).


Xa = 1 & Xb = 0
Xa = 2 & Xb = 1
Xa = 2 & Xb = 0
Xa = 3 & Xb = 2
Xa = 3 & Xb = 1
Xa = 3 & Xb = 0
.
.
.

Do you see the pattern continue?

Then find probability of each case and add them up.
For example:
P( Xa=1 & Xb=0) = P(Xa=1) * P(Xb =0) , and use binomial distribution

Answer 4

typo*
For the first case I got
P( Xa=1 & Xb=0)
= P(Xa=1) * P(Xb =0)
=[ (1200 choose 1 ) * .95^1 * .05^1199 ] * [ (1300 choose 0 ) * .90^0 * .10^1300]

Answer 5

with a calculator you can streamline this a bit.
case 1:
binompdf ( 1200, .95, 1) * binomcdf(1300, .90, 0)

case 2,3
binompdf ( 1200, .95, 2) * binomcdf(1300, .90, 1)

case 4,5
binompdf ( 1200, .95, 3) * binomcdf(1300, .90, 2)


note that im using two different functions
binompdf , and binomcdf

Answer 6

but just to be sure, i could be overcounting, i will manually try this in wolfram

Answer 7

FYI, the answer is according to the text book 0.0104.

Answer 8

typo

I am pretty sure it has to be

binompdf(1200,.95,1200) * binomcdf(1300,.90,1199) +
binompdf(1200,.95,1199) * binomcdf(1300,.90,1198)
+

Answer 9

I used excel and got a sum of 0.010751833

Answer 10

could be because of error , let me try more decimals

Answer 11

what i really want to plug into wolfram is the expression

Sum [binomialpdf(1200,.95,k)*binomialcdf(1300,.90,k-1),{k,1,1200}]

but it does not recognize it.

Answer 12

I used a statistical program called R and got same answer
@dan815

Answer 13

@Zarkon Hi. Can you look this over.
I'm not sure why this approach does not come out to the book answer 0.0104. I also tried using a normal standard curve to approximate binomcdf , but that made it worse.
The statistical package R got the same result 0.010751833

Thanks in advance .