There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are

Question

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. \(\frac{14}{15}\)

M11-08

A.  \frac{8}{15}

B.  \frac{9}{15}

C.  \frac{10}{15}

D.  \frac{12}{15}

E.  \frac{14}{15}

M11-08

Bunuel · Accepted Answer

Official Solution:

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. \(\frac{14}{15}\)

M11-08

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

The only  pair  with a difference greater than 3 is {2, 6}:  (6 - 2 = 4) > 3 . The probability of drawing this pair is  \frac{1}{C^2_6} .
 
Thus, the probability that the absolute difference between the numbers on the cards is 3 or less is:  P = 1 - \frac{1}{C^2_6}= \frac{14}{15} .

Answer: E

Archit3110 · Answer

total possible combinations; 6c2 ; 15
only 1 case where the ∆ will be >3 ; 6-2 ; 4
so we have 14/15 cases where the probability that the positive difference between the numbers on these cards is 3 or less
IMO E

ScottTargetTestPrep · Answer

We see that there are 6C2 = (6 x 5)/2 = 15 ways to choose two numbers from six numbers, and the only way that the difference between two chosen numbers is not 3 or less (i.e., a difference of more than 3) is if the two chosen numbers are 2 and 6. Therefore, there are 14 ways that the difference is 3 or less, so the probability is 14/15.

Answer: E

beeblebrox · Answer

This is how I solved it:

for positive difference to be 0 we will draw the cards like this:
(5,5)
for positive difference to be 1 we will draw the cards like this:
(6,5),(5,6),(5,4),(4,5)
for positive difference to be 2 we will draw the cards like this:
(2,4),(4,2),(6,4),(4,6)
for positive difference to be 3 we will draw the cards like this:
(5,2),(2,5)

==11 ways
Total ways of selecting any two cards:(5,5),(5,2),(2,5), (2,4),(4,2),(6,4),(4,6), (6,5),(5,6),(5,4),(4,5), (6,2), (2,6) ==13 ways

So the ans== 11/13

What did I do wrong? I literally thought that this is the only way we can pick cards without replacement. Also, since we are not replacing the cards, I thought we need to ensure that order matters.

ThatDudeKnows · Answer

beeblebrox

I'll try to explain your mistake by first asking a question. If the deck instead included only 2, 4, 5, 5, 6 (missing one of the 5s), you would have ended up with the same result by deploying your method, right? Does having three 5s instead of two 5s matter? Of course it does! So your method can't be the right one. Why? When you listed (6,5), for example, you really needed to list (6,5a), (6,5b),(6,5c). The same is true for each time you listed a 5.

Okay, so how do we solve it?

How many total ways are there? We have 6 options for which card we draw first. We have 5 options for which card we draw second. So there are 6*5 = 30 total possible ordered pairs. Of those, the only ways to "lose" are (6,2) and (2,6). So 28/30 "win." That's 14/15.

Answer choice E.

missionmba2025 · Answer

Hi  Bunuel

Can you please let me know why are we taking  C^2_6  when three numbers 5, 5, 5 repeat in the set.

For example if we were to find the number of ways one can select two digits from 2, 4, 5, 5, 5, and 6 would we use the following approach

1) Both the numbers don't repeat - Number of ways two numbers can be selected from 2, 4, 5, 5, 5, and 6 such that both the numbers do not repeat =  C^2_4  = 6
2) The numbers repeat - Number of ways two numbers can be selected from 2, 4, 5, 5, 5, and 6 such that both the numbers repeat =  1  = 1

Out of the seven selections made, only one selection in which both 2 and 6 were selected has absolute difference between the numbers on these cards > 3.

So shouldn't the probability = 5/7 ?

Please let me know what I am doing incorrect.

Bunuel · Answer

We are selecting 2 cards out of a set of 6 cards, regardless of the numbers on them, the total number of ways to make this selection is 6C2. This formula represents the number of ways to choose 2 items from a set of 6 items, without considering the order of selection. In this context, the specific numbers on the cards are not relevant to the calculation. We are simply interested in the number of ways to choose any 2 cards from the given set of 6.

We can select the following pairs of cards:

2, 4
2, 5_1
2, 5_2
2, 5_3
2, 6 
4, 5_1
4, 5_2
4, 5_3
4, 6
5_1, 5_2
5_1, 5_3, 
5_1, 6
5_2, 5_3
5_2, 6
5_3, 6

SamAn12213 · Answer

Let the first number be A and the second be B , then |A-B|<=3,,,, -3<=A-B<=3,,,, the difference should be within the specified range,,, only 2 and 6 are not playing by the rules,,, hence 6C2-2C2/6C2=14/15

jnxci · Answer

Such a great question! Thank you for posting it. I have a question concerning a hypothetical variable in the same situation: What IF the card is picked one by one? Would you say that the total number of cases would be 6*5 = 30 instead of 15? And also how can we recognize order or without order better while solving for these type of counting question? Thank you for your reply

Bunuel · Answer

If you choose to count 30 ordered pairs, that’s fine, but you must also double the unfavorable cases. For example, (2, 6) and (6, 2) would both be counted, so the probability stays the same. Either way, the final answer does not change.

amitarya · Answer

I have tried to solved this using traditional probability method.

Please if anyone can held if this method is correct..?

Given Nos. 2 4 5 5 5 6

Method to get atmost abs diff. 3
 
   2, 4 or 4, 2 
 Probability of selecting 2 out of 6 cards * Probability of selecting 4 out of remaining 5 cards*2
 
ð (1/6*1/5*2)

= 2/30 
 
  2,5 or 5,2 
 (1/6*3/5*2)  
 = 6/30 
 
  4,5 or 5,4 
 (1/6*3/5*2) 
 = 6/30 
 
  4,6 or 6.4 
 (1/6*1/5*2) 
 = 2/30 
 
  5,5  
 3/6 * 2/5  
 = 6/30 
 
  5, 6 or 6,5 
 3/6* 1/5 *2 
 = 6/30

Probability of all exclusive events = sum of all = 28/30 = 14/15
Please help, if above method is conceptually correct...

Bunuel

@KarishmaB

Bunuel · Answer

Yes, that's correct. However, as you can see in the official solution, it could have been solved in a simpler way.

BinodBhai · Answer

Are we not considering (2,6) and (6,2) as two different pairs?  Bunuel

Bunuel · Answer

Please review the thread. Your doubt is addressed above.

There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are

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2, 4 or 4, 2	Probability of selecting 2 out of 6 cards * Probability of selecting 4 out of remaining 5 cards2 ð (1/61/5*2)	= 2/30
2,5 or 5,2	(1/63/52)	= 6/30
4,5 or 5,4	(1/63/52)	= 6/30
4,6 or 6.4	(1/61/52)	= 2/30
5,5	3/6 * 2/5	= 6/30
5, 6 or 6,5	3/6* 1/5 *2	= 6/30

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There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are

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