If three fair dice are tossed, what is the probability that

Question

If three fair dice are tossed, what is the probability that exactly one of the numbers that turn up is halfway between the other two?

(A)  \frac{1}{54} 
(B)  \frac{1}{18} 
(C)  \frac{1}{12} 
(D)  \frac{1}{9} 
(E)  \frac{1}{6}

harshshah0079 · Answer

there are total 6 trio satisfying the given condition of exactly one number halfway between other two-(1,2,3),(2,3,4),(3,4,5),(4,5,6),(1,3,5),(2,4,6) -each trio can be arrange by 3!=6 ways
so, favourable outcome =6x6
total outcomes for three dice tossing =6x6x6
so probability=6x6/6x6x6 =1/6

HarshavardhanR · Answer

Total # Outcomes  = 6 x 6 x 6

Favorable Outcomes:

There are only a few cases that will be so such that one of the numbers is halfway between the other 2. Let's count!

Case 1: Spacing of 1 between numbers

The numbers picked could be ---  (1,2,3) OR (2,3,4) OR (3,4,5) OR (4,5,6)

#ways of picking the lowest number = 4C1 ways to pick one number from 1, 2, 3, or 4.
#ways of picking the middle number = 1 (Only 1 option. If 2 is the lowest number, the middle number is 3 and 3 only).
#ways of picking the highest number = 1 (Only 1 option. If 2 is the lowest number, the highest number is 4 and 4 only)

#ways to select the 3 numbers = 4C1 x 1 x 1 = 4

But these numbers could be picked in any order --- (2,3,4), (4,2,3), etc.,...So, we need to also count the #arrangements of the 3 numbers. This would be 3! (because these are 3 different numbers).

#ways of actually getting 3 numbers with spacing of 1 between each = 4 x 3! = 4 x 6

Case 1: Spacing of 2 between numbers
  
 The numbers picked could be ---  (1,3,5) OR (2,4,6)
 
 #ways of picking the lowest number = 2C1 ways to pick one number from 1, 2
#ways of picking the middle number = 1 (Only 1 option. If 2 is the lowest number, the middle number is 4 and 4 only).
#ways of picking the highest number = 1 (Only 1 option. If 2 is the lowest number, the highest number is 6 and 6 only).
#arrangements of the 3 numbers = 3!

#ways of actually getting 3 numbers with spacing of 2 between each = 2 x 3! = 2 x 6

There are no other cases possible.  A spacing of 3, for instance, is impossible.

#Favorable Outcomes = (4 x 6) + (2 x 6) = 6 x 6

The required Probability =  (6 x 6) / (6 x 6 x 6) = 1/6 .

Choice E.

---
Harsha

If three fair dice are tossed, what is the probability that

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