# Probability of Shirts...

Printable View

• April 19th 2010, 12:33 PM
Warrenx
Probability of Shirts...
The soccer team's shirts have arrived in a big box, and people just start grabbing them, looking for the right size. The box contains 4 medium, 10 large, and 6 extra-large shirts. You want a medium for you and one for your sister. Find the probability of the event(s) described below:

d) At least one of the first four shirts you check is a medium.

*I have done this problem at least 5 times, and every time I keeping seeing a different way to look at it*

Here are my rationales:

1) Since they are asking for the chance of getting one out of 4, it would be the bare minimum chance of getting it, so it would be X,X,X, Y. Doing that you get 4/17 or ~23%. (I do not know exactly if this violates the "at least" rule, since it makes sense that you are going for the least common amount).

2) This was my friends method, seems like permutation to me:
1/5, 0, 0 0 = 20%
4/5, 4/19, 0, 0 = (4/5)*(4/19) = 16.8%
4/5, 15/19, 4/18, 0 = (4/5)*(15/19)*(4/18) = 14.3%
4/5, 15/19, 14/18, 4/17 = (4/5)*(15/19)*(14/18)*(4/17) = 11.5%
20 + 16.8 + 14.3 + 11.5 = 62.6%

3) I honestly cannot remember the other 3 ways I was doing it, but one was like 86% so I know that is wrong.

I am doing this purely as guesto-mechanics, and I have no idea what I am looking at or doing, so please if you get the answer try to explain in common sense what is going on so I can get it (Giggle).

Thanks fellas,

Warren.
• April 19th 2010, 02:08 PM
Archie Meade
Quote:

Originally Posted by Warrenx
The soccer team's shirts have arrived in a big box, and people just start grabbing them, looking for the right size. The box contains 4 medium, 10 large, and 6 extra-large shirts. You want a medium for you and one for your sister. Find the probability of the event(s) described below:

d) At least one of the first four shirts you check is a medium.

*I have done this problem at least 5 times, and every time I keeping seeing a different way to look at it*

Here are my rationales:

1) Since they are asking for the chance of getting one out of 4, it would be the bare minimum chance of getting it, so it would be X,X,X, Y. Doing that you get 4/17 or ~23%. (I do not know exactly if this violates the "at least" rule, since it makes sense that you are going for the least common amount).

2) This was my friends method, seems like permutation to me:
1/5, 0, 0 0 = 20%
4/5, 4/19, 0, 0 = (4/5)*(4/19) = 16.8%
4/5, 15/19, 4/18, 0 = (4/5)*(15/19)*(4/18) = 14.3%
4/5, 15/19, 14/18, 4/17 = (4/5)*(15/19)*(14/18)*(4/17) = 11.5%
20 + 16.8 + 14.3 + 11.5 = 62.6%

3) I honestly cannot remember the other 3 ways I was doing it, but one was like 86% so I know that is wrong.

I am doing this purely as guesto-mechanics, and I have no idea what I am looking at or doing, so please if you get the answer try to explain in common sense what is going on so I can get it (Giggle).

Thanks fellas,

Warren.

Hi Warren,

there is a long and short way to answer this.
The most laborious way is as follows...

If one of the first 4 shirts is a medium, then....

A. the first one is a medium and the other 3 are not

the probability of this is $\frac{4}{20}\frac{16}{19}\frac{15}{18}\frac{14}{17 }$

B. the 2nd one is medium and the other 3 are not

the probability is $\frac{16}{20}\frac{4}{19}\frac{15}{18}\frac{14}{17 }$

which is in fact the same as A.

then there is the probability that the 3rd one is medium and the other 3 are not....
and there is the probability that the first 3 are not medium and the 4th one is.

Having done that, you now need to work out the probabilty of 2 of the 4 being medium,
the probability of 3 of the 4 being medium,
the probability of all four being medium.

There is a shortcut... thankfully.

Since all probabilities sum to one,
it's only necessary to calculate the probability of all 4 shirts not being medium
and subtract the answer from 1.

Therefore the probability that at least one shirt is a medium is

$1-P(all\ 4\ are\ not\ medium)=1-\frac{16}{20}\frac{15}{19}\frac{14}{18}\frac{13}{1 7}=1-0.375645=0.624355$

Your friend calculated the following...

If at least 1 medium shirt is picked with 4 attempts,

then there is a $\frac{4}{20}$ probability of picking it at the first try.
Should this happen, we don't need to care what happens next.

There is a $\frac{16}{20}\frac{4}{19}$ probability

that the first medium is chosen at the 2nd attempt, we don't need to care what happens next.

There is a $\frac{16}{20}\frac{15}{19}\frac{4}{18}$

probability of picking the 1st medium shirt on the 3rd go.

There is a $\frac{16}{20}\frac{15}{19}\frac{14}{18}\frac{4}{17 }$

probability the 1st medium shirt is chosen on the 4th try.

Sum those probabilities....
That is also correct!

The answer is 62.435500516%
• April 19th 2010, 05:54 PM
Warrenx
Quote:

Originally Posted by Archie Meade
Hi Warren,

there is a long and short way to answer this.
The most laborious way is as follows...

If one of the first 4 shirts is a medium, then....

A. the first one is a medium and the other 3 are not

the probability of this is $\frac{4}{20}\frac{16}{19}\frac{15}{18}\frac{14}{17 }$

B. the 2nd one is medium and the other 3 are not

the probability is $\frac{16}{20}\frac{4}{19}\frac{15}{18}\frac{14}{17 }$

which is in fact the same as A.

then there is the probability that the 3rd one is medium and the other 3 are not....
and there is the probability that the first 3 are not medium and the 4th one is.

Having done that, you now need to work out the probabilty of 2 of the 4 being medium,
the probability of 3 of the 4 being medium,
the probability of all four being medium.

There is a shortcut... thankfully.

Since all probabilities sum to one,
it's only necessary to calculate the probability of all 4 shirts not being medium
and subtract the answer from 1.

Therefore the probability that at least one shirt is a medium is

$1-P(all\ 4\ are\ not\ medium)=1-\frac{16}{20}\frac{15}{19}\frac{14}{18}\frac{13}{1 7}=1-0.375645=0.624355$

Your friend calculated the following...

If at least 1 medium shirt is picked with 4 attempts,

then there is a $\frac{4}{20}$ probability of picking it at the first try.
Should this happen, we don't need to care what happens next.

There is a $\frac{16}{20}\frac{4}{19}$ probability

that the first medium is chosen at the 2nd attempt, we don't need to care what happens next.

There is a $\frac{16}{20}\frac{15}{19}\frac{4}{18}$

probability of picking the 1st medium shirt on the 3rd go.

There is a $\frac{16}{20}\frac{15}{19}\frac{14}{18}\frac{4}{17 }$

probability the 1st medium shirt is chosen on the 4th try.

Sum those probabilities....
That is also correct!

The answer is 62.435500516%

Thank you for your detailed response Archie Meade. Where can I get reasoning like yours lol! I assume I had read the question wrong then, hehe :).
• April 20th 2010, 02:20 AM
downthesun01
This may be out of the scope of your class but you can use the hypergeometric probability mass function to solve this problem quite simply:

Total number of shirts=N=20, total number of shirts you check=n=4, total number of medium size shirts=k=4, X= the number of medium shirts you find

You're looking for the probability that at least 1 for the four shirts you looked at is size medium.

So, Pr(X $\geq$1)
But this is the same as 1-Pr(X<1), and because X can't be negative, this can be rewritten as 1-Pr(X=0)

So using the hypergeometric PMF:

$1-\frac{(_{x}^{k})(_{n-x}^{N-k})}{(^{N}_{n})}$

$1-\frac{(_{0}^{4})(_{4-0}^{20-4})}{(^{20}_{4})}$

$1-\frac{(_{0}^{4})(_{4}^{16})}{(^{20}_{4})}\approx 0.6244$
• April 20th 2010, 02:58 AM
Warrenx
Quote:

Originally Posted by downthesun01
This may be out of the scope of your class but you can use the hypergeometric probability mass function to solve this problem quite simply:

Total number of shirts=N=20, total number of shirts you check=n=4, total number of medium size shirts=k=4, X= the number of medium shirts you find

You're looking for the probability that at least 1 for the four shirts you looked at is size medium.

So, Pr(X $\geq$1)
But this is the same as 1-Pr(X<1), and because X can't be negative, this can be rewritten as 1-Pr(X=0)

So using the hypergeometric PMF:

$1-\frac{(_{x}^{k})(_{n-x}^{N-k})}{(^{N}_{n})}$

$1-\frac{(_{0}^{4})(_{4-0}^{20-4})}{(^{20}_{4})}$

$1-\frac{(_{0}^{4})(_{4}^{16})}{(^{20}_{4})}\approx 0.6244$

Haha thats an awesome formula Sun. Geez I feel like a toddler now lol :P. Never even considered matrices for this problem, let alone stats.