Roll the dice thrice 2

Probability Level 3

A fair dice is numbered with 1, -2, 0, -1, 3, 2 on its faces.

Let the probability of getting a sum of 5 after rolling the dice thrice can be expressed as $\frac{a}{b}$ where $a,b$ are relatively coprime positive integers.

Find $a+b$ .

Try these too: Part I , Part III .

Image Credit: Flickr Gard Rimestad .

The answer is 77.

1 solution

Ravi Dwivedi
Jul 10, 2015

Total number of cases = $6^3=216$

Sum of 5 can be obtained in following cases

$(3,2,0)$ In this cases there are $3!=6$ ways

$(3,3,-1)$ which can be obtained in $\frac{3!}{2!} = 3$ ways

$(2,2,1)$ in $\frac{3!}{2!} = 3$ ways

$(1,1,3)$ in $\frac{3!}{2!} = 3$ ways

Total number of favourable cases= $6+3+3+3=15$

Required probability = $\frac{15}{216}= \boxed{ \frac{5}{72}}$

Moderator note:

How do we know for certain that we have found all of the "sum of 5"?

Because positive maximum can be 6 so -2 will not be used and 0 can be in only one case when other two are 3 and 2. Remaining cases are intuitively evident by above reasoning.

Ravi Dwivedi - 5 years, 11 months ago

Why $3$ is the exponent?

There is $( 2, 3 )$ , and $( 3, 2 )$ , so only 2 ways to get $5$ , so $\frac { 2 } { 36 } = \frac { 1 } { 18 } = a + b = 1 + 18 = 19.$

And what is the fraction of factorial?

. . - 3 months, 3 weeks ago