Question

If two dice are thrown what are the varrious total numbers of dots that may turns up? What are the probability of each of them? What is the probability that the number of dots will total at least four?

Answer 1

the various total numbers of dots that may turns up: $2-12$

1,1 = 2

1,2 & 2,1 = 3

1,3 & 2,2 & 3,1 = 4

1,4 & 2,3 & 3,2 & 4,1 = 5

1,5 & 2,4 & 3,3 & 4,2 & 5,1 = 6

1,6 & 2,5 & 3,4 & 4,3 & 5,2 & 6,1 = 7

2,6 & 3,5 & 4,4 & 5,3 & 6,2 = 8

3,6 & 4,5 & 5,4 & 6,3 = 9

4,6 & 5,5 & 6,4 = 10

5,6 & 6,5 = 11

6,6 = 12

number of total ways which dots that may turns up: 36

the probability of numbers of dots that may turns up:

$P(2)=1/36,P(3)=2/36=1/18,P(4)=3/36=1/12$

$P(5)=4/36=1/9,P(6)=5/36,P(7)=6/36=1/6$

$P(8)=5/36,P(9)=4/36=1/9,P(10)=3/36=1/12$

$P(11)=2/36=1/18,P(12)=1/36$

the probability that the number of dots will total at least four:

$P(x\geq4)=P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)+P(11)+P(12)$

$P(x\geq4)=\frac{3+4+5+6+5+4+3+2+1}{36}=\frac{33}{36}=\frac{11}{12}$