A bag contains five red marbles and two blue marbles. Quinn draws two marbles at random from the bag without replacement. Afterwards, they put any blue marbles they drew back into the bag, and then redraw marbles until they once again have two marbles. What is the probability they now have two red marbles?

Guest Aug 10, 2023

#1**0 **

Quinn draws two marbles from the bag without replacement. The bag initially contains 5 red marbles and 2 blue marbles.

The probability of drawing a red marble followed by another red marble is given by:

Probability(RR) = (Number of ways to draw 2 red marbles) / (Total number of ways to draw 2 marbles)

Number of ways to draw 2 red marbles = 5C2 (combinations of 5 red marbles taken 2 at a time) Total number of ways to draw 2 marbles = (5 + 2)C2 (total marbles taken 2 at a time)

So, Probability(RR) = 5C2 / (7C2)

victormccormicke Aug 10, 2023

#2**0 **

This yields a probability of (5/7) * (2/6) + (2/7) * (5/6) = 20/42. After replacing the blue marble, the probability of drawing two red marbles becomes (5/7) * (4/6) = 20/42. The final probability of drawing two red marbles in the second round is 20/42 * 20/42 = 400/1764. Simplified, this is 25/147, which is approximately 0.1701, or about 17.01%

Charlieturnbull Aug 10, 2023