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# Five cards are drawn at random from a pack of cards that have been numbered consecutively from 1 to 97, and thoroughly shuffled.

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Five cards are drawn at random from a pack of cards that have been numbered consecutively from 1 to 97, and thoroughly shuffled.

What is the probability that the numbers on the cards, as drawn, are in increasing order of magnitude ?

Guest Dec 7, 2015

#6
+78719
+10

Consider that  all the possible arrangements that can be made by choosing any 5 numbers from 97 = 97*96*95*94*93  = P(97,5)

Now ....consider all the possible sets that can be made by choosing any 5 cards from 97  =

C(97,5)........but .......since these sets can be arranged in any manner.....just order each one from the least number element to the greatest......and we will still have the same number of sets.....

So....the probability is given by :

[Number of ordered arrangements ]   /   [ Number of possible arrangements]  =

C(97,5)  / P(97/5)   =  (   97! /  [ (97 - 5)! *5! ] )  / [97! / (97 - 5)!]  =

[97! / 97!] * [ 97 - 5]!/ [97 - 5]! * [ 1 / 5! ]  = 1/120   =  about .833.....%

CPhill  Dec 8, 2015
edited by CPhill  Dec 8, 2015
Sort:

#2
+1035
+5

There are (5!) possible arrangements of the 5 cards; one arrangement is (completely) in increasing order of magnitude.

The probability is

$$\Large {\frac{1}{5!}} \small \text{ or } \Large {\frac {1}{120}}\\$$

Nauseated  Dec 8, 2015
#3
+26329
+10

Consider a simpler problem first

Alan  Dec 8, 2015
#4
+5

Actually though it looks like Alan has made a mistake in his arithmetic right at the end.

nCr(97,5)/nPr(97,5) = 1/5! = 1/120 = 0.00833... .

Interestingly, both answers show that the number of cards in the pack is irrelevant, (so long as it is five or more).

Guest Dec 8, 2015
#5
+26329
+5

"Actually though it looks like Alan has made a mistake in his arithmetic right at the end."

I did indeed make a mistake in the arithmetic of the last line.  It should have been:

nCr(97,5)/nPr(97,5) = 0.00833...   (= 1/5!)

In general:   nCr(n,k)/nPr(n,k) = n!/[(n-k)!k!]*(n-k)!/n!  = 1/k!   (as given directly by Nauseated).

Alan  Dec 8, 2015
#6
+78719
+10

Consider that  all the possible arrangements that can be made by choosing any 5 numbers from 97 = 97*96*95*94*93  = P(97,5)

Now ....consider all the possible sets that can be made by choosing any 5 cards from 97  =

C(97,5)........but .......since these sets can be arranged in any manner.....just order each one from the least number element to the greatest......and we will still have the same number of sets.....

So....the probability is given by :

[Number of ordered arrangements ]   /   [ Number of possible arrangements]  =

C(97,5)  / P(97/5)   =  (   97! /  [ (97 - 5)! *5! ] )  / [97! / (97 - 5)!]  =

[97! / 97!] * [ 97 - 5]!/ [97 - 5]! * [ 1 / 5! ]  = 1/120   =  about .833.....%

CPhill  Dec 8, 2015
edited by CPhill  Dec 8, 2015
#7
+91038
+5

I really like your explanation Chris :)

Melody  Dec 8, 2015

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