+0

# Interesting probability question on Mathcounts

0
471
5
+2665

4 numbers are randomly selected from the subset \((1, 2,3,4,5,6,7,8,9)\). What is the probability that the product of the 4 randomly selected integers is a multiple of 14?

I got 4/81, but I have doubts about it, and I can't settle it.

Note: This is not the original question, but it is enough from my memory to solve the problem, and I can post this because it has been over 24 hours since I took the Mathcounts, and you can no longer take it.

Feb 19, 2022
edited by BuilderBoi  Feb 19, 2022

#1
+118571
-1

Deleted becasue it was not correct

Feb 19, 2022
edited by Melody  Feb 19, 2022
#2
+2

Melody: The computer lists 52 such numbers as follows:

1237 , 1247 , 1257 , 1267 , 1278 , 1279 , 1347 , 1367 , 1378 , 1457 , 1467 , 1478 , 1479 , 1567 , 1578 , 1678 , 1679 , 1789 , 2347 , 2357 , 2367 , 2378 , 2379 , 2457 , 2467 , 2478 , 2479 , 2567 , 2578 , 2579 , 2678 , 2679 , 2789 , 3457 , 3467 , 3478 , 3479 , 3567 , 3578 , 3678 , 3679 , 3789 , 4567 , 4578 , 4579 , 4678 , 4679 , 4789 , 5678 , 5679 , 5789 , 6789 , Total = 52 such numbers.

Probability ==52 / 126 ==26 / 63

Feb 19, 2022
#3
+118571
+3

Thanks guest.  You are totally correct.

I failed to recognise that all the even numbers have a factor of 2

I'll start again

4 numbers are randomly selected from the subset (1,2,3,4,5,6,7,8,9) . What is the probability that the product of the 4 randomly selected integers is a multiple of 14?

One of the numbers has to be 7.  AND at least one has to be even.

Forget about the 7 and there are 4 even and 4 odd numbers from which to choose the other numbers

3 evens                                  4C3                       = 4

2 evens and an odd                4C2*4C1 = 6*4   = 24

1 even and 2 odds                 4C1* 4C2 = 4*6   = 24

Total  4+24+24 = 52               (just like guest said)

Prob = 52 / 4C9 =  52 / 126

Now we are in agreement.

Melody  Feb 19, 2022
edited by Melody  Feb 19, 2022
#4
+2665
-1

Thanks Guest and Melody!!

I put 4/81 because there was \(9^4\) ways to choose the 4 numbers

I thought there was only 4 x 1 x 9 x 9 options, because 1 has to be 7, 1 has to be of the 4 even numbers listed in the subset, and the remaining 2 don't matter.

I think I accidentally accounted for order

Thank's again both Melody and Guest!!!

BuilderBoi  Feb 19, 2022
#5
+118571
+3

You have made 2 mian logic errors here.

1) You have said that all the numbers can be used more than once, which they can't.

2) You also said that order chosen counts, you already understand that you made that mistake.

Learning is a journey :)

Melody  Feb 19, 2022