Stumped: Four military recruits whose respective shoe sizes are 7,8,9 & 10....

[RobertMcCadden]

I am studying for the PE exam and working my way through some probability questions and this one has got me stumped:

"Four military recruits whose respective shoe sizes are 7,8,9 & 10 report to the supply clerk to be issued boots. The supply clerk selects one pair of boots in each of the four required sizes and hands them at random to the recruits. What is the probability that all recruits will receive boots of an incorrect size?"

I know the total number of combinations is 4!=24, but I cannot determine how many incorrect combinations resulting in all recruits receiving the incorrect size there are without writing out all the combinations. If I look at each recruit separately the probability of the first recruit receiving the incorrect size is 3/4, but then things fall apart when I look at the second recruit because the first recruit may or may not have gotten the second recruits shoe size.

Anybody know the answer to this?

 

[tkhunny]

I am studying for the PE exam and working my way through some probability questions and this one has got me stumped:

"Four military recruits whose respective shoe sizes are 7,8,9 & 10 report to the supply clerk to be issued boots. The supply clerk selects one pair of boots in each of the four required sizes and hands them at random to the recruits. What is the probability that all recruits will receive boots of an incorrect size?"

I know the total number of combinations is 4!=24, but I cannot determine how many incorrect combinations resulting in all recruits receiving the incorrect size there are without writing out all the combinations. If I look at each recruit separately the probability of the first recruit receiving the incorrect size is 3/4, but then things fall apart when I look at the second recruit because the first recruit may or may not have gotten the second recruits shoe size.

Anybody know the answer to this?

What is the probability that all recruits will receive boots of an incorrect size?"

Sometimes it helps to rethink...

What is the probability that NO recruit will receive boots of an correct size?"

Sometimes, with a short list, you can just write it all down.

789T - 7 is right - No good.
78T9 - 7 is right - no good.
798T - 7 is right - no good
79T8 - 7 is right - no good
7T89 - 7 is right - no good
7T98 - 7 is right - no good
879T - 9 is right - no good.
87T9
897T - 10 is right - no good.
89T7
8T79
8T97 - 9 is right - no good.
978T - 10 is right - no good.
97T8
987T - 8 is right - no good.
98T7 - 8 is right - no good.
9T78
9T87
T789
T798 - 9 is right - no good.
T879 - 8 is right - no good.
T897 - 8 is right - no good.
T978
T987

There are other ways to go about it. On a high-pressure exam, speed may be important.

 

[RobertMcCadden]

What is the probability that all recruits will receive boots of an incorrect size?"

Sometimes it helps to rethink...

What is the probability that NO recruit will receive boots of an correct size?"

Sometimes, with a short list, you can just write it all down.

789T - 7 is right - No good.
78T9 - 7 is right - no good.
798T - 7 is right - no good
79T8 - 7 is right - no good
7T89 - 7 is right - no good
7T98 - 7 is right - no good
879T - 9 is right - no good.
87T9
897T - 10 is right - no good.
89T7
8T79
8T97 - 9 is right - no good.
978T - 10 is right - no good.
97T8
987T - 8 is right - no good.
98T7 - 8 is right - no good.
9T78
9T87
T789
T798 - 9 is right - no good.
T879 - 8 is right - no good.
T897 - 8 is right - no good.
T978
T987

There are other ways to go about it. On a high-pressure exam, speed may be important.

Thank you for your reply tkhunny, I actually ended up writing similar to how you did; however, I though there may be an equation I could use to get the answer and that I was possibly missing an important concept. Would you mind elaborating a little on the other ways to go about solving this problem?

Thanks again

 

[Dr.Peterson]

I am studying for the PE exam and working my way through some probability questions and this one has got me stumped:

"Four military recruits whose respective shoe sizes are 7,8,9 & 10 report to the supply clerk to be issued boots. The supply clerk selects one pair of boots in each of the four required sizes and hands them at random to the recruits. What is the probability that all recruits will receive boots of an incorrect size?"

I know the total number of combinations is 4!=24, but I cannot determine how many incorrect combinations resulting in all recruits receiving the incorrect size there are without writing out all the combinations. If I look at each recruit separately the probability of the first recruit receiving the incorrect size is 3/4, but then things fall apart when I look at the second recruit because the first recruit may or may not have gotten the second recruits shoe size.

Anybody know the answer to this?

There is actually a formula for this, which you can read about here, or here.

I can't quite imagine giving this as an exam question and expecting you to figure all that out on the spot; is it possible you are expected to know the formula?

 

[pka]

There is actually a formula for this, which you can read about here, or here.
I can't quite imagine giving this as an exam question and expecting you to figure all that out on the spot; is it possible you are expected to know the formula?

Re: Not giving this as an exam question. In my graduate course on counting theory & probability this topic was one my favorite.
I like as this a beautiful interplay between \(\displaystyle \displaystyle e =\sum\limits_{k = 0}^\infty {\dfrac{1}{{k!}}} \) and the count of derangements.
If \(\displaystyle \mathscr{D}(n)\) denotes the number of derangements of \(\displaystyle n\) items then \(\displaystyle \mathscr{D}(n)=\left\lfloor {\dfrac{{n!}}{e} + 0.5} \right\rfloor \). It is easy to see that \(\displaystyle \mathscr{D}(4)=9\).
It is reasonable to expect anyone to know that formula for a test.

 

[JeffM]

I cannot remember the formula, and it is a tricky counting problem.

How many ways can you get all four correct. Obviously, just 1.

How many ways can you get exactly three correct? ZERO. If one soldier gets the wrong boots, then that soldier has got the boots of a different soldier, who must necessarily also get the wrong boots.

Now one pair of soldiers could each get the other's boots and a different pair of soldiers each get the right boots. So exactly two soldiers can get the correct boots. How many ways? Well that is obviously

\(\displaystyle \dbinom{4}{2} = 6.\)

Can exactly one soldier get the correct size of boots? The answer is yes, but the number of ways that can happen is not necessarily obvious. Figure that out, and you can figure out how many ways no soldier gets the right size of boots.

 

[pka]

I cannot remember the formula, and it is a tricky counting problem.
How many ways can you get all four correct. Obviously, just 1.

I disagree that the formula is at all trickey.
Because the derangement of four items is nine; there is \(\displaystyle 1-\dfrac{9}{4!}\) probability that at least one person will get the correct size boot.

 

[mmm4444bot]

I disagree that the formula is at all trickey.

Who said the formula is trickey?

 

[pka]

Who said the formula is trickey?

I cannot remember the formula, and it is a tricky counting problem.

above

 

[mmm4444bot]

I cannot remember the formula, and it is a tricky counting problem.

above

There's no need to shout.

I think you misread Jeff's statement. He referred to the counting problem in the op as tricky, not the formula. (Note the comma after 'formula' -- it indicates a compound sentence.)