Rolling Three Dice

The number of ways to have all dice showing the same number is $4$ (they all show $1, 2, 3, 4$ ). The total number of outcomes of the dice rolls is $4 \cdot 6 \cdot 10 = 240$ . Hence $P(all the same) = \frac{4}{240} = \frac{1}{60}$ Therefore $a + b = 1 + 60 = \fbox{61}$

There are $4\cdot6\cdot10=240$ possible outcomes. The dice only show the same number when they all show $1$ , $2$ , $3$ , or $4$ , for a total of $4$ ways. The probability of them all showing the same number is then $\frac{4}{240}=\frac{1}{60}$ . Adding $1$ and $60$ gives us $\boxed{61}$ .

There are only 4 ways for all 3 die to show the same number, which is if they show the numbers 1/2/3/4.

In order to calculate the total number of possible outcomes, which is 4 x 6 x 10 =240 outcomes, we need to understand that for every 1 of the 4 numbers displayed on the 4-sided dice, there will be 6 possible numbers for the second dice and 10 for the third dice. Therefore, you should get 240 ways in total.

Thus, 4/240 = 1/60. Sum is 61.

assume that 4 dice.........showed some number .........for others to show same number probability=1/6*1/10=1/60 so=60+1=61

Sides that are common = 4 $$ Total Combinations of 4,6 & 10 sided dice are $$ $4 \times 6 \times 10 = 240$

Dividing $\frac{4}{240} =\frac{1}{60} = 1+60 = \boxed{61}$

The number of ways of rolling the three dice is $\large 4 \times 6 \times 10 =240$ . Although there is only $\large 4$ numbers that all three dice share. Hence the probability that they all roll the same number is $\large \frac{4}{240}=\frac{1}{60}$ . Therefore $\large a+b=1+60=\boxed{61}$ .