Answer:
[tex]\dfrac{2}{9}[/tex]
11 times
Step-by-step explanation:
When you roll two number dice, you can get 36 different outcomes:
[tex]\begin{array}{cccccc}(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)&(14,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}[/tex]
The sum of 7 give outcomes:
[tex](1,6),\ (2,5),\ (3,4),\ (4,3),\ (5,2),\ (6,1)[/tex]
- 6 outcomes in total.
The sum of 11 give outcomes:
[tex](5,6),\ (6,5)[/tex]
- 2 outcomes in total.
So, the probability that a roll of two dice will produce a sum of either 7 or 11 is
[tex]\dfrac{6+2}{36}=\dfrac{8}{36}=\dfrac{2}{9}[/tex]
In 50 trials, you can expect
[tex]\dfrac{2}{9}\cdot 50=\dfrac{100}{9}=11\dfrac{1}{9}\approx 11[/tex]
times this event to occur
