+2668
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.
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
+118670 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.
+2668 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!!!
