Two different digits are chosen at random from the set 1,2,3,4,5,6,7,8

Question

Two different digits are chosen at random from the set  . What is the probability that their sum will exceed 13?

A.  \frac{1}{7}

B.  \frac{1}{12}

C.  \frac{1}{14}

D.  \frac{5}{14}

E.  \frac{3}{14}

rajatchopra1994 · Answer

Explanation:

Selecting 2 number from set = 8C2 = 8x7/2 = 28

Pair which has sum greater than 13, (8,7 or 8,6) = 2

Probability = 2/28 = 1/14

IMO-C

HarshaBujji · Answer

It's always better to manually count the n.o sets that satisfy the condition rather than using a formula in the case of small n.os.

As sum is 13 > (8,7) , (8,6) are only subsets that are possible out of 8C2 = 28;

So p(n) = 1/14;

IMO C

Kinshook · Answer

Given: Two different digits are chosen at random from the set  .

Asked: What is the probability that their sum will exceed 13?

Digits 1, 2, 3, 4, 5 cannot be selected since 8 + 5 = 13 and 2 digits sum does not exceed 13.

Possible combinations = {(6,8),(7,8)} : 2 ways
Total ways = 8C2 = 28

The probability that their sum will exceed 13 = 2/28 = 1/14

IMO C

Archit3110 · Answer

Total ways to select 2 digits 
8c2=28
Pair of numbers sum>13 (6,8;7,8)
2
P =2/28;1/14
Option C

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MANASH94 · Answer

Can someone help: 
Why are we not considering (8,7 and 7,8) as two different cases? Similarly for (6,8 and 8,6)

Bunuel · Answer

Those are indeed two different cases, and if you use the probability approach, the answer would be (1/8 * 1/7 * 2) + (1/8 * 1/7 * 2). This gives 1/14, which is correct. The factor of 2 in each case accounts for the two possible sequences of drawing: 7 then 8, and 8 then 7; 6 then 8, and 8 then 6.

However, when you solve using the combination approach, the denominator is 8C2, which counts  unordered  pairs of two numbers out of eight. In the numerator, you also have just two  unordered  cases: {7,8} and {6,8}. Since both numerator and denominator consistently disregard order, you again get the correct answer.

So, the key here to maintain consistency between the numerator and denominator.

Hope it's clear.

viswanath1 · Answer

Why are we not choosing 7,8 and 8,7 like we do in dice problems

MANASH94 · Answer

A way for explanation might be that when we choose two random numbers together then it will not matter in which order you get the numbers. Like in Dice problems if you throw dice one after the another it would matter but if you throw two identical dice in one go it would not matter.
So with the two numbers, if choose two numbers in one go it will not matter.

A better response has been posted earlier by  Bunuel . You may refer that as well.

Two different digits are chosen at random from the set 1,2,3,4,5,6,7,8

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