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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 difference between the numbers on these cards is 3 or less?

I solved this question in this way but got answer wrong What is wrong with my method?

Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - probability to select 2 and 6=1-1/6*1/6=35/36

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 difference between the numbers on these cards is 3 or less?

I solved this question in this way but got answer wrong What is wrong with my method?

Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - probability to select 2 and 6=1-1/6*1/6=35/36

The probability of chosing 2 and 6 =1/6 each

Probability of selecting two cards from (2,6) = 2C2 / 6C2 = 2/15 Hence, the required probability = 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 difference between the numbers on these cards is 3 or less?

I solved this question in this way but got answer wrong What is wrong with my method?

Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - probability to select 2 and 6=1-1/6*1/6=35/36

The probability of chosing 2 and 6 =1/6 each

Seems the question is little ambigious: how the difference is calculated? Is it x-y or y-x? x = first y = last

When 2 and 6 are slelcted, 6-2 = 4 but 2-6 = -4. Seems in this case order matters because if 2 comes first and 6 comes last, it should be 2-6 and vice versa.

If I am correct, the only difference that is greater than 3 is 6-2 = 4. In that case, then the prob is 1 - 1/(6P2) = 1 - 1/30 = 29/30.

I differ with most of the answer and even few explanation...below is my answer and explanation....correct me, if anything

Approach 1: Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - 1/6C2 = 1 - 1/ 15 = 14/15 (as there is only one such combination that has diff more than 3 i.e. (2,4))

Approach 2: How many such combinations are there which has diff of <= 3...there are 14...you need to manually find out...(2,4), (2,5), (2,5), (2,5), (4,5), etc.. So, Probability that the difference is 3 or less = 14/ 6C2 = 14/15

Obviously Approach 1 is much faster...choosing right approach with some quick intelligence saves lot of time in exam....

I differ with most of the answer and even few explanation...below is my answer and explanation....correct me, if anything

Approach 1: Probability that the difference is 3 or less = 1 - probability that the difference is more than 3 = 1 - 1/6C2 = 1 - 1/ 15 = 14/15 (as there is only one such combination that has diff more than 3 i.e. (2,4))

Approach 2: How many such combinations are there which has diff of <= 3...there are 14...you need to manually find out...(2,4), (2,5), (2,5), (2,5), (4,5), etc.. So, Probability that the difference is 3 or less = 14/ 6C2 = 14/15

Obviously Approach 1 is much faster...choosing right approach with some quick intelligence saves lot of time in exam....

I have question regarding approach 1.

If (2,4) = (4,2) and is considered only 1 combination...then the total combination is:

According to my calculations I receive 17/18 =(34/36) which is different from 14/15

Please help why I am wrong?

pkit here the total number of possibilities are not 36 but 30.( you have considered 6 sets which are (2,2),(4,4),(5,5),(6,6).. and these sets shouldn't be considered)

so a total of 30 sets of values are present. There is one and only one possibility that the difference is greater than 3 that is the set (6,2) we cannot consider (2,6) bcause the value becomes a negative and less than 3.