Dice O' Dice!

Rolling a die will have 6 possible outcomes {1, 2, 3, 4, 5, and 6}. By Multiplication Rule of the Counting Principle , the number of elements in the sample space in rolling three dice is

$n(S)= 6\times 6\times 6 =216$

The probability of getting at least two 4's is the probability of getting two 4's or getting three 4's. These two events have no elements in common.

A- be the event of rolling two 4's. B- the event of rolling three 4's.

Event A: The list of outcomes of getting two 4's in rolling three dice: {(1 4 4), (4 1 4), (4 4 1), (2 4 4), (4 2 4), (4 4 2), (3 4 4), (4 3 4), (4 4 3), (5 4 4), (4 5 4), (4 4 5), (6 4 4), (4 6 4), (4 4 6)}

$n(A)=15$

$P(A)= \frac{15}{16}$

Event B: There is only one way of getting three 4's in rolling three dice, that is {(4 4 4)}

$P(B) = \frac{1}{16}$

Therefore, the probability of getting at least 2 4's is:

$\frac{15}{216} + \frac{1}{216} = \frac{2}{27}$

Hence, $a=2$ and $b=27$ . $a+b= \boxed{29}$

getting two 4s on the 1st two rolls = (1/6)(1/6)

the third roll can have any number because of the condition so the probability is now

(1x1x6)/(6x6x6)

now we consider the 3 variances of the dice results with the 4s being on the 1st and 2nd roll, 1st and 3rd, 2nd and 3rd: (1x1x6x3)/(6x6x6)

and also remove from these 3 variances where the result is all 4 (4,4,4) which occurs only once, so we subtract 2, removing the same occurences:

[(1x1x6x3)-2]/(6x6x6)

simplifying (18-2)/216 ==> 16/216 ==> 2/27 2+27 = 29