probability that the maximum and median occurs exactly once in set

[MaryStew]

I was wondering if someone could confirm my answer for the 1st question and then maybe guide me how to do the 2nd question?


Let n be an odd integer. The \emph{median} of a sequence x1,…,xn, denoted of numbers, median(x1,…,xn), is the unique value x such that there are at least ⌈n/2⌉ values less than or equal to x and at least ⌈n/2⌉ values greater than or equal to x.

For example,

median(8,4,3,4,7,5,6)=5 since 4,3,4,5≤5 and 8,7,5,6≥5; and

median(8,5,3,4,7,5,6)=5 since 5,3,4,5≤5 and 8,5,7,5,6≥5.


Consider a uniform random sequence x1,…,x7 of 7 numbers each chosen from the set {1,2,3,…,10}


a. What is the probability that max{x1,…,x7} occurs exactly once in x1,…,x7?

I did:

n=7, that is 7*(1⁶+2⁶+3⁶+4⁶+5⁶+6⁶+7⁶+8⁶+9⁶) / 10⁷ = .6848835

b. What is the probability that median(x1,…,x7) occurs exactly once in x1,…,x7?

I'm a little confused with this one. My friends are getting around 0.47 but I don't understand how. I would appreciate any help.

 

[Romsek]

[MATH]\text{Suppose $k$ is the median}\\ \text{Then the 3 sequence members above the median must be $(k+1),\dots 10$}\\ \text{And the 3 members below must be $1,\dots (k-1)$}\\ p = \dfrac{7\cdot \dbinom{6}{3} \sum \limits_{k=2}^9 (k-1)^3 (10-k)^3}{10^7} = \dfrac{7476}{15625} \approx 0.478464 [/MATH]

 

[Steven G]

[MATH]\text{Suppose $k$ is the median}\\ \text{Then the 3 sequence members above the median must be $(k+1),\dots 10$}\\ \text{And the 3 members below must be $1,\dots (k-1)$}\\ p = \dfrac{7\cdot \dbinom{6}{3} \sum \limits_{k=2}^9 (k-1)^3 (10-k)^3}{10^7} = \dfrac{7476}{15625} \approx 0.478464 [/MATH]

It seems that you are saying that 1 and 10 must be included in the 7 numbers. Why?

EDIT: OK, maybe you are saying that the numbers must be chosen from (k+1) to 10 and ....
Now my new question is why can't 'k-1' and 'k+1" be equal?

 

[Romsek]

So now my question is why can't k-1 and k+1 actually be the same number? .

The median occurs once.

 

[Steven G]

It seems that you are saying that 1 and 10 must be included in the 7 numbers. Why?

The median occurs once.

So you did part b.

 

[Romsek]

So you did part b.

well yeah, OP correctly did part a.
part a referred to the max of the sequence
part b refers to the median

the line \(\displaystyle \text{Suppose $k$ is the median}\) should be a bit of a giveaway

maybe I see the confusion.

The sequence elements may be chosen from 1...(k-1), and (k+1)...10
They don't have to be any one of those in particular, but those two sets are what the members below and above the median can be drawn from.

 

[Romsek]

[MATH]\text{Suppose $k$ is the median}\\ \text{Then the 3 sequence members above the median must be $(k+1),\dots 10$}\\ \text{And the 3 members below must be $1,\dots (k-1)$}\\ p = \dfrac{7\cdot \dbinom{6}{3} \sum \limits_{k=2}^9 (k-1)^3 (10-k)^3}{10^7} = \dfrac{7476}{15625} \approx 0.478464 [/MATH]

This is poor terminology.

When I say that the 3 sequence members above the median must be \(\displaystyle (k+1),\dots, 10\) I mean that they are drawn from that set.
Same with the 3 below the median. Those elements must come from the set \(\displaystyle 1,\dots, (k-1)\)

 

[bobMarle]

a. What is the probability that max{x1,…,x7} occurs exactly once in x1,…,x7?

I did:

n=7, that is 7*(1⁶+2⁶+3⁶+4⁶+5⁶+6⁶+7⁶+8⁶+9⁶) / 10⁷ = .6848835


MaryStew, why did you make the equation 1⁶, 2⁶.. ?

 

[Dr.Peterson]

a. What is the probability that max{x1,…,x7} occurs exactly once in x1,…,x7?
I did:
n=7, that is 7*(1⁶+2⁶+3⁶+4⁶+5⁶+6⁶+7⁶+8⁶+9⁶) / 10⁷ = .6848835

MaryStew, why did you make the equation 1⁶, 2⁶.. ?

Do you see that the denominator is the number of possible sequences?

The numerator has to be the number of such sequences in which the maximum occurs exactly once.

How many ways can the maximum be 1, and occur exactly once?

How many ways can the maximum be 2, and occur exactly once?

Keep going, and you should see the reasoning. If you don't, then let us know how you are thinking, so we can guide you further.