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siddhantvarma
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M11-08 12 Apr 2025, 00:03
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Bunuel KarishmaB These cards have to be considered distinct because it isn't explicitly stated in the question that they're identical right?
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M11-08 12 Apr 2025, 01:09
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siddhantvarma

There are six cards with numbers 2, 4, 5, 5, 5, and 6 on them. If two cards are randomly selected without replacement, what is the probability that the absolute difference between the numbers on these cards is at most 3?

Bunuel KarishmaB These cards have to be considered distinct because it isn't explicitly stated in the question that they're identical right?
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Yes, here all 6 cards are considered distinct — even though some have the same number written on them.
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M11-08 14 Jun 2025, 04:22
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I like the solution - it’s helpful.
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M11-08 19 Jul 2025, 23:08
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I like the solution - it’s helpful.
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isteducimus
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M11-08 16 Sep 2025, 13:29
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Here we have considered all 5 as different as they were given to be cards. If these had been simply a set of numbers (and not card), would prob have been 9/15 in that case?
Bunuel

There are THREE cards with 5: ­2, 4, 5', 5'', 5''', and 6.

(2, 4)
(2, 5')
(2, 5'')
(2, 5''')
(2, 6)

(4, 5')
(4, 5'')
(4, 5''')
(4, 6)

(5', 5'')
(5', 5''')
(5', 6)

(5'', 5''')
(5'', 6)

(5''', 6)
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M11-08 16 Sep 2025, 13:45
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isteducimus
Here we have considered all 5 as different as they were given to be cards. If these had been simply a set of numbers (and not card), would prob have been 9/15 in that case?

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No, it would not be 9/15.

If you treat it as just a set of distinct numbers {2, 4, 5, 6}, then there are only C(4,2) = 6 pairs total. The only pair with difference greater than 3 is {2,6}, so the favorable ones are 5.

So in that case the probability would be 5/6, not 9/15.
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M11-08 17 Sep 2025, 10:37
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Grateful for the response. if the set is {2,4,5,5,5,6} - total cases would be 6C2/3!?
probability would be 1 - 1/total cases = 1-(2/5)=3/5=9/15
is the understanding correct?

Bunuel


No, it would not be 9/15.

If you treat it as just a set of distinct numbers {2, 4, 5, 6}, then there are only C(4,2) = 6 pairs total. The only pair with difference greater than 3 is {2,6}, so the favorable ones are 5.

So in that case the probability would be 5/6, not 9/15.
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M11-08 17 Sep 2025, 11:05
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isteducimus
Grateful for the response. if the set is {2,4,5,5,5,6} - total cases would be 6C2/3!?
probability would be 1 - 1/total cases = 1-(2/5)=3/5=9/15
is the understanding correct?


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6C2 = 15
3! = 6

15/6 = 5/2. How can the number of total cases be 2.5?

Not sure what you’re doing there, but it’s definitely not correct. So I’ll just say no.
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