If three different integers are selected at random from the integers 1 through 8, what is the probability...

[BTGmoderatorLU]

Source: Manhattan Prep

If three different integers are selected at random from the integers 1 through 8, what is the probability that the three selected integers can be the side lengths of a triangle?

A. 11/28
B. 27/56
C. 1/2
D. 4/7
E. 5/8

The OA is A

[Jay@ManhattanReview]

Source: Manhattan Prep

If three different integers are selected at random from the integers 1 through 8, what is the probability that the three selected integers can be the side lengths of a triangle?

A. 11/28
B. 27/56
C. 1/2
D. 4/7
E. 5/8

The OA is A

Note that one of the important properties of a triangle is that the sum of any two sides must be greater than the third side.

Keeping that in mind, let's pick and choose the integers that follow the above rule. The eligible values are

• (2, 3, 4) (3, 4, 5) (4, 5, 6) (5, 6, 7) (6, 7, 8): No. of ways = 5
• (2, 4, 5) (3, 4, 6) (4, 5, 7) (5, 6, 8): No. of ways = 4
• (2, 5, 6) (3, 5, 6) (4, 5, 8) (5, 7, 8): No. of ways = 4
• (2, 6, 7) (3, 5, 7) (4, 6, 7): No. of ways = 3
• (2, 7, 8) (3, 6, 7) (4, 6, 8): No. of ways = 3
• (3, 6, 8) (4, 7, 8): No. of ways = 2
• (3, 7, 8): No. of ways = 1

Total no. of ways = 22

Required probability = 22/56 = 11/28

The correct answer: A

Hope this helps!

-Jay
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[Scott@TargetTestPrep]

Source: Manhattan Prep

If three different integers are selected at random from the integers 1 through 8, what is the probability that the three selected integers can be the side lengths of a triangle?

A. 11/28
B. 27/56
C. 1/2
D. 4/7
E. 5/8

The OA is A

The total number of ways to randomly pick 3 integers from 8 (without any restrictions) is:

8C3 = (8 x 7 x 6)/(3 x 2) = 8 x 7 = 56

To ensure the 3 integers picked can be the side lengths of a triangle, the sum of the two smaller integers must be greater than the largest integer. Let’s list all the triples that have this property (notice that the smallest integer that can be in a triple is 2):

(2, 3, 4), (2, 4, 5), (2, 5, 6), (2, 6, 7), (2, 7, 8)

(3, 4, 5), (3, 4, 6), (3, 5, 6), (3, 5, 7), (3, 6, 7), (3, 6, 8), (3, 7, 8)

(4, 5, 6), (4, 5, 7), (4, 5, 8), (4, 6, 7), (4, 6, 8), (4, 7, 8)

(5, 6, 7), (5, 6, 8), (5, 7, 8)

(6, 7, 8)

We see that there are 5 + 7 + 6 + 3 + 1 = 22 such triples. Therefore, the probability is 22/56 = 11/28,

Answer: A