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There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?

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

There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?

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

The only pair which has the positive difference more than 3 is {2,6}: \(6-2=4 \gt 3\). Probability of this pair to be drawn is \(\frac{1}{C^2_6}\).

So, the probability that the positive difference between the numbers on the cards is 3 or less is: \(P=1-\frac{1}{C^2_6}=\frac{14}{15}\).

Doesn't the use of the phrase "positive difference between the numbers on these cards" mean that we need to eliminate the possibility of picking two 5s as well. Difference between two 5s is 0, which is technically not a difference at all.

There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?

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

The only pair which has the positive difference more than 3 is {2,6}: \(6-2=4 \gt 3\). Probability of this pair to be drawn is \(\frac{1}{C^2_6}\).

So, the probability that the positive difference between the numbers on the cards is 3 or less is: \(P=1-\frac{1}{C^2_6}=\frac{14}{15}\).

Answer: E

Hi Bunnel,

6C2 is number of ways in which cards can be chosen, not pairs.

there are 2 ways in which this particular pair of number can be chosen (6,2 or 2,6)

So shouldn't it be 1 - 2/15 = 13/15??
_________________

There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?

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

The only pair which has the positive difference more than 3 is {2,6}: \(6-2=4 \gt 3\). Probability of this pair to be drawn is \(\frac{1}{C^2_6}\).

So, the probability that the positive difference between the numbers on the cards is 3 or less is: \(P=1-\frac{1}{C^2_6}=\frac{14}{15}\).

Answer: E

Hi Bunnel,

6C2 is number of ways in which cards can be chosen, not pairs. Choosing randomly 2 cards from the set means choosing pairs

there are 2 ways in which this particular pair of number can be chosen (6,2 or 2,6) (6,2) and (2,6) are considered the same since we dont cover the order in counting

There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?

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

The only pair which has the positive difference more than 3 is {2,6}: \(6-2=4 \gt 3\). Probability of this pair to be drawn is \(\frac{1}{C^2_6}\).

So, the probability that the positive difference between the numbers on the cards is 3 or less is: \(P=1-\frac{1}{C^2_6}=\frac{14}{15}\).

Answer: E

Bunuel, would you please elaborate more on your answer? I initially approached this via 6!/3! to get rid of the duplicates and this gives 120, but I believe this is wrong because the problem implies that ordering does not matter because the problem states 'positive difference'. However I do not understand 6C2. What about the duplicate 5s?

A simpler example would be a list of numbers 2,3,3 and I am trying to choose two numbers. 3C2 would give me (2,3), (3,3) and (2,3) for 3 combinations, but should it not only be 2?

There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?

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

I have a couple of doubts while solving this question . 1. if its not given the cards are distinct ,do we assume they are? 2. while eliminating the number pairs why do we not consider 6,2?