Question

If you roll a pair of fair dice, what is the probability of each of the following? (round all answers to 4 decimal places, .XXXX)
a) getting a sum of 1?
b) getting a sum of 5?
c) getting a sum of 12?

Answer 1

Answer:  The required probabilities are

$(a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.$

Step-by-step explanation:  Given that a pair of fair dice is rolled.

We are to find the probability of getting

(a) getting a sum of 1.

b) getting a sum of 5.

c) getting a sum of 12.

Let S be the sample space for the experiment of rolling a pair of fair dice.

Then, S = {(1,1), (1,2), (1,3), (1, 4), (1,5), (1,6), .  . . , (6,5), (6,6)}.

And, n(S) =36.

(a) Let E denote the event of getting a sum of 1.

Since the sum of the numbers on two dice is minimum 2, so

E = { }  ⇒  n(E) = 0.

Therefore, the probability of event E is

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{0}{36}=0.$

(b) Let F denote the event of getting a sum of 5.

Then,

F = {(1,4), (2,3), (3,2), (4,1)}  ⇒  n(F) = 4.

Therefore, the probability of event F is

$P(F)=\dfrac{n(F)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9}.$

(c) Let G denote the event of getting a sum of 12.

Then,

G = {(6,6)}  ⇒  n(G) = 1.

Therefore, the probability of event G is

$P(G)=\dfrac{n(G)}{n(S)}=\dfrac{1}{36}.$

Thus, the required probabilities are

$(a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.$