Question

two numbered 1 to 6 dice are thrown together and their scores are added. the probability that the sum will be 4 is?​

Answer 1

Answer:

$Pr = \frac{3}{36}$

Step-by-step explanation:

Given

2 number die

Required

P(Sum = 4)

The sample space of 2 die is:

$S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)$

$(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6), (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)$

$(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6), (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)$

$n(S)= 36$

The pairs that adds up to 4 are:

$Sum(4) = \{(1,3), (2,2),(3,1)\}$

$n(Sum) =3$

So, the probability is:

$Pr = \frac{n(Sum)}{n(S)}$

$Pr = \frac{3}{36}$