Answer: [tex]\dfrac{2}{9}[/tex]
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 = [tex]n(S)=6\times6=36[/tex]
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 :-
[tex]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}[/tex]
Hence, the required probability = [tex]\dfrac{2}{9}[/tex]
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
