<u>Answer-</u>
<em>The probability of winning on the first roll is </em><em>0.22</em>
<u>Solution-</u>
As in the game of casino, two dice are rolled simultaneously.
So the sample space would be,
![|S|=6^2=36](https://tex.z-dn.net/?f=%7CS%7C%3D6%5E2%3D36)
Let E be the event such that the sum of two numbers are 7, so
E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
![|E|=56](https://tex.z-dn.net/?f=%7CE%7C%3D56)
![\therefore P(E)=\dfrac{|E|}{|S|}=\dfrac{6}{36}](https://tex.z-dn.net/?f=%5Ctherefore%20P%28E%29%3D%5Cdfrac%7B%7CE%7C%7D%7B%7CS%7C%7D%3D%5Cdfrac%7B6%7D%7B36%7D)
Let F be the event such that the sum of two numbers are 11, so
F = {(6,5), (5,6)}
![|F|=2](https://tex.z-dn.net/?f=%7CF%7C%3D2)
![\therefore P(F)=\dfrac{|F|}{|S|}=\dfrac{2}{36}](https://tex.z-dn.net/?f=%5Ctherefore%20P%28F%29%3D%5Cdfrac%7B%7CF%7C%7D%7B%7CS%7C%7D%3D%5Cdfrac%7B2%7D%7B36%7D)
Now,
![P(\text{sum is 7 or 11)}=P(E\ \cup\ F)=P(E)+P(F)=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}=0.22](https://tex.z-dn.net/?f=P%28%5Ctext%7Bsum%20is%207%20or%2011%29%7D%3DP%28E%5C%20%5Ccup%5C%20F%29%3DP%28E%29%2BP%28F%29%3D%5Cdfrac%7B6%7D%7B36%7D%2B%5Cdfrac%7B2%7D%7B36%7D%3D%5Cdfrac%7B8%7D%7B36%7D%3D%5Cdfrac%7B2%7D%7B9%7D%3D0.22)