In a game, a player draws a card from a stack three of cards showing the numbers 3, 4, and 5, and then the player draws a card from a second stack that contains 5 cards showing the numbers 1, 2, 3, 4, and 5. which table shows the probability distribution of the sum of two drawn cards?

How I got this

We can eliminate choices A and C because an outcome of "1" is not possible. This is because the smallest outcome allowed is 4, from adding 3 and 1. If we were just focused on drawing 1 card from the second stack, then an outcome of 1 is possible.

We can also eliminate choice B because the probabilities do not add to 1. Add up all the fractions shown in table B and you'll get the following:

1/15 + 2/15 + 1/5 + 4/15 + 1/5 + 2/15 + 1/15 = 1.067 approximately

All of the probabilities must add up to 1 for a probability distribution to be possible. This is why choice C is eliminated.

The only thing left is choice D

Add up the probabilities in choice D

1/15 + 2/15 + 1/5 + 1/5 + 1/5 + 2/15 + 1/15 = 1

we get the proper result of 1

Each probability in this table is found by dividing the number of times the outcome shows up out of 15. So for example, the outcome of "4" only happens one time out of 15 total, which is why 1/15 is the probability for this outcome. The fraction 1/5 is equivalent to 3/15.

answer: plz put a pic in so i can answer it

step-by-step explanation: