written 8.3 years ago by
teamques10
★ 66k

•
modified 8.3 years ago

When 2 fair dice are thrown, the sample space is given by
$S= $$
\begin{Bmatrix}
\
(1,1) & (1,2) & (1,3)&(1,4)&(1,5)&(1,6) \\
\ (2,1) & (2,2) & (2,3)&(2,4)& (2,5)&(2,6)\\
(3,1) & (3,2) & (3,3) &(3,4)&(3,5)&(3,6) \\
\ (4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\
\ (5,1) &(5,2)& (5,3)&(5,4)&(5,5)&(5,6) \\
\ (6,1) &(6,2)&(6,3)&(6,4)&(6,5)&(6,6) \\
\end{Bmatrix} $
$$n(S)=36$$
Probability mass function (pmf) for the sum of the faces is given by
![enter image description here][1]
Cumulative distribution function (cdf) for the sum of the faces is given by
![enter image description here][2]
Let ‘A’ be the event that the product of the faces is 12
A={(2,6),(3,4),(4,3),(6,2)}
n(A)=4
$$∴P(A)=n(A)/n(S) =4/36=1/9$$
∴ The probability that the product of the faces is 12 = 1/9