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 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 and let P(E) be the probability of event E, then When x = 4, then P(E) is equal toA. `5//9`B. `6//7`C. `7//18`D. `8//19`

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 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 and let P(E) be the probability of event E, then When x = 4, then P(E) is equal toA. `5//9`B. `6//7`C. `7//18`D. `8//19`
Answer» Correct Answer - C The number of ways in which different sums can occurs is (3 + 2 + 1)(1 + 2 + 3) = 36. The probability of 4 is `14//36 = 7//18`.

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