Answer:
0.2
Step-by-step explanation:
If two number cubes as tossed then the total possible out comes are
Total = {(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)}
Number of elements in sample space are
n(S)=36
Let A and B represent the following events.
A = The sum will be 4.
A = {(1,3),(2,2),(3,1)}
n(A) = 3
B = The sum is less than or equal to 6.
B = {(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)}
n(B)=15
Intersection of both events is
A∩B = {(1,3),(2,2),(3,1)}
n(A∩B) = 3
![P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{36}](https://tex.z-dn.net/?f=P%28A%29%3D%5Cdfrac%7Bn%28A%29%7D%7Bn%28S%29%7D%3D%5Cdfrac%7B3%7D%7B36%7D)
![P(B)=\dfrac{n(B)}{n(S)}=\dfrac{15}{36}](https://tex.z-dn.net/?f=P%28B%29%3D%5Cdfrac%7Bn%28B%29%7D%7Bn%28S%29%7D%3D%5Cdfrac%7B15%7D%7B36%7D)
![P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{3}{36}](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3D%5Cdfrac%7Bn%28A%5Ccap%20B%29%7D%7Bn%28S%29%7D%3D%5Cdfrac%7B3%7D%7B36%7D)
We need to find the probability that the sum will be 4, given that the sum is less than or equal to 6.
![P(\frac{A}{B})=\dfrac{P(A\cap B)}{P(B)}](https://tex.z-dn.net/?f=P%28%5Cfrac%7BA%7D%7BB%7D%29%3D%5Cdfrac%7BP%28A%5Ccap%20B%29%7D%7BP%28B%29%7D)
![P(\frac{A}{B})=\dfrac{\frac{3}{36}}{\frac{15}{36}}](https://tex.z-dn.net/?f=P%28%5Cfrac%7BA%7D%7BB%7D%29%3D%5Cdfrac%7B%5Cfrac%7B3%7D%7B36%7D%7D%7B%5Cfrac%7B15%7D%7B36%7D%7D)
![P(\frac{A}{B})=\dfrac{3}{15}](https://tex.z-dn.net/?f=P%28%5Cfrac%7BA%7D%7BB%7D%29%3D%5Cdfrac%7B3%7D%7B15%7D)
![P(\frac{A}{B})=\dfrac{1}{5}](https://tex.z-dn.net/?f=P%28%5Cfrac%7BA%7D%7BB%7D%29%3D%5Cdfrac%7B1%7D%7B5%7D)
![P(\frac{A}{B})=0.2](https://tex.z-dn.net/?f=P%28%5Cfrac%7BA%7D%7BB%7D%29%3D0.2)
Therefore, the required probability is 0.2.