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You shouldn't use the binomial distribution here as it is a multinominal distribution problem (a generalization of the binomial).

So let's gather what we have:

n = 6 (total number of events)
n1 = 2 in part 1 (pudding #1)
n2 = 2 in part 2 (pudding #2)
n3 = 2 in part 3 (pudding #3)
p1 = 2/6 (probability to get 2 from n1)
p2 = 2/6 (probability to get 2 from n2)
p3 = 2/6 (probability to get 2 from n3)

the formula goes like this:

so let's put the numbers in motion:

$p = \frac{6!}{(2!*2!*2!)} *(2/6)^2 * (2/6)^2 * (2/6)^2 = 0.12345$

and we get

10/81

You shouldn't use the binomial distribution here as it is a multinominal distribution problem (a generalization of the binomial).

So let's gather what we have:

n = 6 (total number of events)
n1 = 2 in part 1 (pudding #1)
n2 = 2 in part 2 (pudding #2)
n3 = 2 in part 3 (pudding #3)
p1 = 2/6 (probability to get 2 from n1)
p2 = 2/6 (probability to get 2 from n2)
p3 = 2/6 (probability to get 2 from n3)

the formula goes like this:

so let's put the numbers in motion:

$p = \frac{6!}{(2!*2!*2!)} *(\frac{2}{6})^2 * (\frac{2}{6})^2 * (\frac{2}{6})^2 = 0.12345 $

and we get

10/81

You shouldn't use the binomial distribution here as it is a multinominal distribution problem (a generalization of the binomial).

So let's gather what we have:

n = 6 (total number of events)
n1 = 2 in part 1 (pudding #1)
n2 = 2 in part 2 (pudding #2)
n3 = 2 in part 3 (pudding #3)
p1 = 2/6 (probability to get 2 from n1)
p2 = 2/6 (probability to get 2 from n2)
p3 = 2/6 (probability to get 2 from n3)

the formula goes like this:

so let's put the numbers in motion:

p = (6!/(2!*2!*2!)) *( (2/6)^2 * (2/6)^2 * (2/6)^2 ) = 0.12345  //yes, I need to learn LaTex :-)

and we get

10/81

You shouldn't use the binomial distribution here as it is a multinominal distribution problem (a generalization of the binomial).

So let's gather what we have:

n = 6 (total number of events)
n1 = 2 in part 1 (pudding #1)
n2 = 2 in part 2 (pudding #2)
n3 = 2 in part 3 (pudding #3)
p1 = 2/6 (probability to get 2 from n1)
p2 = 2/6 (probability to get 2 from n2)
p3 = 2/6 (probability to get 2 from n3)

the formula goes like this:

so let's put the numbers in motion:

$p = \frac{6!}{(2!*2!*2!)} *(2/6)^2 * (2/6)^2 * (2/6)^2 = 0.12345$

and we get

10/81

You can't use binomial distribution here as it is a multinominal distribution. So let's gather what we have:

n = 6 
n1 = 2 in part 1 (pudding #1)
n2 = 2 in part 2 (pudding #2)
n3 = 2 in part 3 (pudding #3)
p1 = 2/6 (probability to get 2 from n1)
p2 = 2/6 (probability to get 2 from n2)
p3 = 2/6 (probability to get 2 from n3)

the formula goes like this:

so let's put the numbers in motion:

p = (6!/(2*2*2)) *( (2/6)^2 * (2/6)^2 (2/6)^2 ) = 0.12345  //yes, I need to learn LaTex :-)

and we get

10/81

You shouldn't use the binomial distribution here as it is a multinominal distribution problem (a generalization of the binomial). 

So let's gather what we have:

n = 6 (total number of events)
n1 = 2 in part 1 (pudding #1)
n2 = 2 in part 2 (pudding #2)
n3 = 2 in part 3 (pudding #3)
p1 = 2/6 (probability to get 2 from n1)
p2 = 2/6 (probability to get 2 from n2)
p3 = 2/6 (probability to get 2 from n3)

the formula goes like this:

so let's put the numbers in motion:

p = (6!/(2!*2!*2!)) *( (2/6)^2 * (2/6)^2 * (2/6)^2 ) = 0.12345  //yes, I need to learn LaTex :-)

and we get

10/81