Probablity

• Jun 14th 2008, 05:33 AM
marky001
Probablity
Hi all,
I am new here but am coming to the end of the first year of my course so will probably be posting loads of questions over the next few weeks!! The first on I am struggling with is a probability question.

A package contains 80 similar components and inspection shows that four have been damaged during transit. If six components have been drawn at random from the contents of the package determine the probability that in this sample
a) one and
b) Less than three

are damaged.

I thought that I could solve this by calculating the probability that one component is damaged on a draw and multiplying by six

i.e. 76/80 * 75/79 * 74/78 * 73/77 * 72/76 * 4/75 = 0.041
0.041 * 6 = 0.25

This has been marked wrong. I tried something similar for question b but that was also marked as wrong. Can anyone help??

Mark
• Jun 14th 2008, 05:51 AM
mr fantastic
Quote:

Originally Posted by marky001
Hi all,
I am new here but am coming to the end of the first year of my course so will probably be posting loads of questions over the next few weeks!! The first on I am struggling with is a probability question.

A package contains 80 similar components and inspection shows that four have been damaged during transit. If six components have been drawn at random from the contents of the package determine the probability that in this sample
a) one and
b) Less than three

are damaged.

I thought that I could solve this by calculating the probability that one component is damaged on a draw and multiplying by six

i.e. 76/80 * 75/79 * 74/78 * 73/77 * 72/76 * 4/75 = 0.041
0.041 * 6 = 0.25

This has been marked wrong. I tried something similar for question b but that was also marked as wrong. Can anyone help??

Mark

Have you been taught the hypergeometric distribution?
• Jun 14th 2008, 06:52 AM
marky001
hypergeometric distribution
No. Tis is part of what our lecturer calls 'independent learning'. By this he gives us some questions we know nothing about and a set of notes from a text book. However I cant seem to work this one out and the notes don't mention hypergeometric distribution.

MM
• Jun 14th 2008, 07:01 AM
Moo
Hello,

76 safe, 4 damaged, agree ?

Actually, what I'm going to do is like the hypergeometric distribution. The main difference is that if you know this distribution, you have the direct formula, that's the power of the lazy people ! (Rofl)

You want to pick up 6 components among 80. So that's ${80 \choose 6}$ possible drawings.

Let X be the number of damaged components you pull out from the package.

a) you want only one damaged, $X=1$. So it's like choosing one component among the 4 damaged + choosing 5 components among the 76 which are not damaged.
So ${4 \choose 1} \times {76 \choose 5}$ represents the number of possible drawings containing exactly 1 damaged component.

$P(X=1)=\frac{{4 \choose 1} \times {76 \choose 5}}{{80 \choose 6}} \approx 0.2459$

b) $P(X<3)=P(X=0)+P(X=1)+P(X=2)$

Same reasoning as above =)

Sidenote : ${n \choose k}=\frac{n!}{k!(n-k)!}$ is the number of ways to draw k elements from a set of n elements.
• Jun 15th 2008, 04:42 AM
marky001
Thanks
Many thanks for taking the time to help me out with this! It's really appreciated!

MM