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siddhantvarma

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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?

Bunuel

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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?

Yes, here all 6 cards are considered distinct — even though some have the same number written on them.

bondi123

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14 Jun 2025, 04:22

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I like the solution - it’s helpful.

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19 Jul 2025, 23:08

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I like the solution - it’s helpful.

isteducimus

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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)

Bunuel

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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?

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.

isteducimus

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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

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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?

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.

NikunjC

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09 Dec 2025, 10:48

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I like the solution - it’s helpful. Why can i use this formula to determine the number of ways to select 2 out of 6 when 3 of the values are 5 ?

6!/3! = 20

and then i was getting 19 by 20.

Bunuel

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09 Dec 2025, 11:01

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NikunjC

I like the solution - it’s helpful. Why can i use this formula to determine the number of ways to select 2 out of 6 when 3 of the values are 5 ?

6!/3! = 20

and then i was getting 19 by 20.

Because you are selecting physical cards, not distinct values.

Even if three cards show 5, they are still three different cards. So the total number of equally likely 2-card selections is C(6,2) = 15.

Your 6!/3! = 120 is the wrong tool here anyway. And 6!/3! = 20 is not correct arithmetic.

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08 Feb 2026, 10:16

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Deconstructing the Question
Six cards have numbers 2, 4, 5, 5, 5, 6.
Two cards are selected without replacement.
We want the probability that the absolute difference is at most 3.

Step-by-step
Total number of unordered pairs:
\(C(6,2) = 15\)

Count favorable pairs (|difference| ≤ 3):

(2,4): 1 way
(2,5): 3 ways
(2,6): excluded

(4,5): 3 ways
(4,6): 1 way

(5,5): \(C(3,2)=3\) ways
(5,6): 3 ways

Total favorable outcomes:
\(1+3+3+1+3+3 = 14\)

Probability:
\(\frac{14}{15}\)

Answer: 14/15

Suyog99

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09 Sep 2026, 06:45

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I don’t quite agree with the solution. Its At most 3 , not atleast 3. So this is the wrong solution

Bunuel

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09 Sep 2026, 06:53

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Suyog99

I don’t quite agree with the solution. Its At most 3 , not atleast 3. So this is the wrong solution

The solution is correct. “At most 3” means the difference can be 0, 1, 2, or 3.

So it is easier to exclude the cases where the difference is greater than 3. The only such pair is {2, 6}.

Thus the desired probability is:

1 - 1/15 = 14/15.

Please re-read the question and the solution carefully.

Hope it helps.

Docmba68

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18 Sep 2026, 00:37

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Bunuel

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?

A. \(\frac{8}{15}\)
B. \(\frac{9}{15}\)
C. \(\frac{10}{15}\)
D. \(\frac{12}{15}\)
E. \(\frac{14}{15}\)

Always remember in questions like these where language says “ at least or a most “ take a pause and remind yourself of this:-
P(success)=1- P(failure)

Here, the only way to "fail" (absolute difference > 3) is to draw the pair {2, 6}. The difference is 6 - 2 = 4.

Scenario A: Draw the 2 first, then the 6

· Draw 1: Probability of drawing a 2 = 1/6 (There is one '2' out of six cards).
· Draw 2: Probability of drawing a 6 = 1/5 (One card is gone, so 5 remain. There is one '6' left).
· Probability of Scenario A: 1/6 * 1/5 =1/30

Scenario B: Draw the 6 first, then the 2

· Draw 1: Probability of drawing a 6 = 1/6 (There is one '6' out of six cards).
· Draw 2: Probability of drawing a 2 = 1/5 (One card is gone, so 5 remain. There is one '2' left).
· Probability of Scenario B: 1/6* 1/5 =1/30

Total Probability of Failure:
Add the two scenarios together:
P(failure) = 1/30 + 1/30 =1/15

Now P(success)= 1-1/15 =14/15

Answer option(E)

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