M11-08

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}

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}

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

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

SchruteDwight · Answer

The probability approach is as follows.

6,2 can be arranged in 2 ways

1-2*\frac{1}{6}*\frac{1}{5}

axezcole · Answer

Total conditions-6p2
Condition that satisfy difference greater than 3= (2,6) (6,2)

Prob=1-2/30

Prob=14/15

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

einstein801 · Answer

Shouldnt the denominator be 6C1*5C1 since there's no replacement?

Bunuel · Answer

No replacement implies that we get two different cards, which is exactly what 6C2 = 15 gives: the number of combinations of two different cards from 6. In contrast, 6C1 * 5C1 = 30 gives the number of permutations of two different cards, where the order matters. So, in this case, you get both (1, 2) and (2, 1), hence twice the number compared to 6C2.

Vikramaditya00 · Answer

I think this is a  high-quality  question and I agree with explanation. My logic may be flawed but can do it directly like only one pair exceeds the difference that is (6,2), so if we look at the options see that there is only one option E where the difference between the numerator and denominator is of 1, so can we directly choose that option?

Bunuel · Answer

How would you have chosen if there was another option with the difference between the numerator and the denominator being 1?

Shadyshades · Answer

Bunuel

I could count only 7 total possible outcomes. Can you please correct me here ?

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

Bunuel · Answer

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)

Shadyshades · Answer

Ah ! Totally missed that they are cards. It could be 5 of spade,5 of diamond,5 of heart or 5 of club. Hence, they shouldn't be considered same outcome.
Thank you.

karansinghania · Answer

Although, I got the same anwer, just want to understand any specifc reason why you have not considered all possible "arrangements" here. While I understand that if we multiple the numerator and denominator with 2! to account for those cases, it eventually cancels out. But logic wise using combinations is somehow not making sense to me. Can you please give some clarity to avoid confusion in other questions?

Bunuel · Answer

The question refers to the absolute difference between the numbers. The absolute difference between, say, 3 and 5 is the same as the absolute difference between 5 and 3. Therefore, there's no need to account for different arrangements in the pairs of numbers.

karansinghania · Answer

In that case, probability of selecting the pair {2, 6} would be 2/6 * 1/5, right? Where 2/6 represents probability of drawing either a 2 or 6 and 1/5 represents probability of drawing the remaining number. This includes both cases {2,6} and {6,2}.

But when we do it using the formulae it is 1/6C2, which automatically multiplies the number by 2!. Just finding this logic tricky.

Bunuel · Answer

6C2 gives pairs without order. Just check the pairs listed above and find out which satisfy the condition, then calculate the probability to double-check. Here is another discussion on this question for your reference: https://gmatclub.com/forum/there-are-si ... 14375.html Hlope it helps.

kll08 · Answer

Hi  Bunuel   KarishmaB ,

I have a basic doubt. Even if I use the FCP(Fundamental Counting Principle) to arrive at the total possible outcomes i.e. 1st choice can be made from 6 card and second from 5 cards, so total outcomes possible are 5*6 = 30.

As per my understanding,   FCP never takes into account order and gives possible combinations  . So, I can't understand why we considered 15 instead of 30.

Also, do you have a general differentiator that we can use to distinguish whether we need to consider all 3 5's as just 5(duplicates) or each as unique?

Thanks

KarishmaB · Answer

First choice can be made in 6 ways. - In one of these 6 ways, you select a 2. Now second choice can be made in 5 ways. In one of those 5 ways, you select a 6. - In one of these 6 ways, you select a 6. Now second choice can be made in 5 ways. In one of those 5 ways, you select a 2. So of the 30 ways, there are 2 ways in which you select a 2 and a 6. In FCP, the slots are distinct (such as first pick and second pick here). So arrangement is taking place. Here is a video discussing FCP: https://youtu.be/LFnLKx06EMU Also, whether elements are identical or distinct is given to us. In some cases, if it could be either, it is best for us to assume they are distinct. In probability, overall it makes no impact. So say it is about outcomes on a pair of dice. I will assume one die is red and other is yellow if I am not given that they must be identical. Helps avoid errors. But some objects such as fruits and vegetables, blue pencils, red marbles etc are usually considered identical. A physical card with a 5 on it could be of any color. So we assume they are distinct. If we were instead given numbers - of the 6 numbers, 2, 4, 5, 5, 5, 6, two are picked - then picking the two 5s is one case. It cannot be 3 cases.

phmahi1997 · Answer

I like the solution - it’s helpful.

M11-08

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