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28 Oct 2021, 09:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:59) correct
65%
(02:48)
wrong
based on 106
sessions
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There are two decks of 10 cards each. The cards in each deck are labeled with integers from 11 to 20 inclusive. If we pick a card from each deck at random, what is the probability that the product of the numbers on the picked cards is a multiple of 6?
A. 0.23
B. 0.36
C. 0.40
D. 0.42
E. 0.46
A. 0.23
B. 0.36
C. 0.40
D. 0.42
E. 0.46
Updated on: 31 Oct 2021, 17:10
Originally posted by JerryAtDreamScore on 28 Oct 2021, 11:20.
Last edited by JerryAtDreamScore on 31 Oct 2021, 17:10, edited 1 time in total.
Last edited by JerryAtDreamScore on 31 Oct 2021, 17:10, edited 1 time in total.
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Bunuel
Breaking Down the Info:
We have \(10 * 10 = 100\) possible products in total. We need (i) a factor of 6, or (ii) a factor of 2 with a factor of 3 to create a multiple of 6.
(i) We can start from 12 and 18, they can take any number so 12 * X, 18 * X, X * 12, and X * 18 are all easy multiples of 6. That is 40 possible options but remove 4 of them as we double-counted 12*12, 18*18, 12*18, and 18*12, hence 36 options.
(ii) Next, we can try a multiple of 2 times a multiple of 3. The multiples of 2 are {14, 16, 20} (remove 12 and 18 as we used them earlier).
The only multiple of 3 is {15} (again remove 12 and 18 as we used them earlier).
Hence that is only 6 more options if we permutate them and reverse their orders.
Then we have 36 + 6 = 42 options in total.
Answer: D
29 Oct 2021, 00:25
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Quote:
Total number of possible outcomes: 10*10 = 100
There are 4 possible cases:
(1) first card is not multiple of 6, cards are 11, 13, 17, 19; 4 ways. The other card has to be multiple of 6, that is 12, 18; 2 ways. For elaboration these are (11,12), (11,18), (13,12), (13,18), (17,12), (17,18), (19,12), (19,18). There are 8 ways
(2) first card multiple of 6, cards are 12, 18; 2 ways. The other is any card out of 10 cards; 10 ways. There are 2*10 = 20 ways
(3) first card multiple of 2 but not of 3, cards are 14, 16, 20; 3 ways. The other card is multiple of 3, cards are 12, 15, 18; 3 ways. There are 3*3 = 9 ways
(4) first card multiple of 3 but not of 2, cards are 15; 1 ways. The other card is multiple of 2, cards are 12, 14, 16, 18, 20; 5 ways. There are 1*5 = 5 ways
Total number of required outcomes i.e. picking a card which is multiple of 6 are 8+20+9+5 = 42 ways
Required probability = 42/100 = 0.42
D is correct
