Unto zero

Probability Level 3

Let $a, b$ and $c$ be the three numbers resulting from three independent throws of a fair 6-sided dice labeled $1$ through $6$ . The probability that $(a-b)(b-c)=0$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are coprime positive integers. What is the value of $m+n$ ?

The answer is 47.

2 solutions

Vighnesh Raut
Mar 22, 2014

Total outcomes = 6 X 6 X 6 = 216

Now the above equation (a-b)(b-c)=0 is only possible if either of the two (a-b) and (b-c) is zero or both of them are 0.

There are three cases :

Case 1 = a is equal to b. (Choices=6 P 2) There are 6 choices for a and only 1 choice for b as both are equal and 5 choices for c as c is neither equal to a nor to b. Therefore, there are 6 X 1 X 5 =30 choices where a=b.

Case 2= b is equal to c. Then in the same way as above , there are 6 choices for b and only 1 choice for c and 5 choice for a. Therefore , there are again 6 X 1 X 5 = 30 choices where b=c.

Case 3= All [a,b,c] are equal .(choices= 6 P 1) Now, for this to be true there are only 6 choices [ (1,1,1) , (2,2,2) ,.......,(6,6,6) ]

So , total possible outcomes = 30 + 30+ 6 =66 choices .

Therefore, probability = 66 divided by 216 = 11 / 36

So, answer = 11 + 36 = 47.

I came to 49, but i see now that i double counted the possibility of 3 of the same numbers. The double use of a and b in the question might be confusing for some.

Peter van der Linden - 4 years, 7 months ago

Bhaskar Krishnan
Jan 8, 2017

P(A) is a=b which is 6/36 i.e 1/6 similarly P(B) is b=c which is also 1/6. Now P(A intersection B) is a=b=c which is 1/36. Now P(A) or P(B) = P(A) + P(B) - P(A intersection B) Which is 1/6+1/6-1/36 =11/36 So m+n = 47

Unto zero

The answer is 47.

2 solutions

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