A bag contains 5 red, 9 white, and 10 blue marbles. Suppose you choose a marble from the bag and then choose another marble without replacing the first one. Find the probability of picking the same color both times.
= Prob(both red) + Prob(both white) + prob(both blue)
= (5/24)(4/23) + (9/24)(8/23) + (10/24)(9/23)
= 5/138 + 3/23 + 15/92
= 91/276
To find the probability of picking the same color both times, we need to consider the possible outcomes for each pick.
The first pick can result in a red, white, or blue marble. Since there are a total of 5 red, 9 white, and 10 blue marbles, the probability of picking a red on the first pick is 5/(5+9+10) = 5/24. Similarly, the probability of picking a white or blue on the first pick is 9/24 and 10/24 respectively.
After picking the first marble, there will be one less marble of that color in the bag. So for the second pick, the probability of picking the same color as the first pick would be slightly different.
Let's consider the case when the first pick is a red marble. After removing one red marble, there are 4 red, 9 white, and 10 blue marbles left in the bag. Therefore, the probability of picking another red marble from the remaining marbles is 4/(4+9+10) = 4/23.
By following the same reasoning, the probability of picking the same color for the second pick, given that the first pick was white, would be 8/23. Similarly, if the first pick was blue, the probability of picking another blue marble would be 9/23.
Since there are three possible scenarios for the first pick (red, white, or blue), we can calculate the overall probability by multiplying the probability of each scenario.
(5/24) * (4/23) + (9/24) * (8/23) + (10/24) * (9/23) = 20/276 + 72/276 + 90/276 = 182/276 = 91/138 ≈ 0.659
Therefore, the probability of picking the same color both times is approximately 0.659 or 65.9%.