Question

A pair of dice is rolled, and the sum of the numbers is either 7 or 11. What is the probability of this event?

Answer 1

Answer:

Probability that sum of numbers is either 7 or 11 is:

0.22

Step-by-step explanation:

A pair of dice is rolled.

Sample Space:

(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)

Total outcomes= 36

Outcomes with sum of numbers either 7 or 11 are in bold letters=8

i.e. number of favorable outcomes=8

So, P(sum of numbers is either 7 or 11 )=8/36

                                                                   =0.22

Answer 2

Answer:   $\dfrac{2}{9}$

Step-by-step explanation:

Let A be the event that the sum is 7 and and B be the event that the sum is 11 .

When two pair of dices rolled the total number of outcomes = $n(S)=6\times6=36$

The sample space of event A ={(1,6), (6,1), (5,2), (2,5), (4,3), (3,4)}

Thus n(A)= 6

The sample space of event B = {(5,6), (6,5)}

n(B)=2

Since , both the events are independent , then the required probability is given by :-

$P(A\cup B)=P(A)+P(B)\\\\=\dfrac{n(A)}{n(S)}+\dfrac{n(B)}{n(S)}=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}$

Hence, the required probability = $\dfrac{2}{9}$