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(E) be the probability of event E, thenWhen x=4, then P(E) is equal to

There are two die A and B both having six faces. Die A has three faces

1.	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(E) be the probability of event E, thenWhen x=4, then P(E) is equal to
Answer» 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 $P (E)$ be the probability of event $E$ , then When $x = 4$ , then $P (E)$ is equal to