Question

There are two die $A$ and $B$ both having six faces. Die $A$ has three faces marked with $1$, two faces marked with $2$, and one face marked with $3$. Die $B$ has one face marked with $1$, two faces marked with $2$, and three faces marked with $3$. Both dices are thrown randomly once. If $E$ be the event of getting sum of the numbers appearing on top faces equal to $x$, let $P\left(E\right)$ be the probability of event $E$, then
When $x=4$, then $P\left(E\right)$ is equal to

• A. $\frac{5}{9}$
• B. $\frac{6}{7}$
• C. $\frac{7}{18}$
• D. $\frac{8}{19}$

Answer 1

The correct option is C $\frac{7}{18}$

$x$ can be $2,3,4,5,6.$ The number of ways in which sum of $2,3,4,5,6$ can occur by the coefficients of $x^2,x^3,x^4,x^4,x^5,x^6$ in
$(3x+2x^2+x^3)(x+2x^2+3x^3)=3x^2+8x^3+14x^4+8x^5+3x^6$
The number of ways in which different sums can occur is
$(3+2+1)(1+2+3)=36$.
The probability of $4$ is $\frac{14}{36}=\frac{7}{18}$