Question

places and are E and F independent
A box has 10 marbles in it, 4 red and 6 white. Suppose we draw a marble from the box, replace it, and then draw another. Find the probability that
(a) Both marbles are red
(b) Just one of the two marbles is red

Answer 1

(a) The probability that the first marble is red is 4/10. Since you are replacing the marble after you draw it, the probability that the second marble is red is also 4/10. To get the probability that both marbles are red, you multiply 4/10 by 4/10 to get 16/100=4/25.
(b) The probability that just one marble is red is the probability that the 1st marble is red times the probability that the 2nd marble is not red, plus the probability that the 1st marble is not red times the probability that the 2nd marble is red. Thus, you get (4/10)*(6/10) + (6/10)*(4/10), or .48. (Correct me if I'm wrong.)

Answer 2

a. 4/10 * 4/10 There are 4 out of 10 to pick from then you multiply the next draw which would still be 4/10 that are red. So the answer is 16/100 which reduces down to 4/25

b. 4/10 * 6/10 There are 4 out of 10 to pick from that would give you red and then there are 6/10 which would give you white.. Nowyou will have 1 red. So the answer is 24/100 which reduces down to 6/25

Answer 3

melliness...you were right on the dot with the answers. blexting your first answer was correct but now your second I am adding some more questions if you guys could help me out that would be great