Sorry here is completed question
A bag contains 3 red marbles, 3 blue marbles and 3 yellow marbles. Which of the following methods of selecting 2 marbles will have the greater probability of choosing 2 red marbles?
1- Randomly choose a marble don't replace it then randomly choose another marble
2 - Randomly choose a marble replace it then randomly choose another marble
You could have added a posting to the original question. Anyway,
There are 3 red marbles out of 9 total, so on the first draw,
P(red) = 3/9 = 1/3
Without replacement, there are now 2 of 8 reds, so P(red) = 2/8 = 1/4
P(red,red) = 1/3 * 1/4 = 1/12
With replacement, both draws are like the first, so
P(red,red) = 1/3 * 1/3 = 1/9
To determine which method of selecting two marbles will have a greater probability of choosing two red marbles, we need to calculate the probabilities for each method.
Method 1: Randomly choose a marble without replacement, then randomly choose another marble.
Since there are 3 red marbles in the bag and we are choosing without replacement, the probability of choosing a red marble on the first pick is 3/9. After the first red marble is chosen, there are 2 red marbles left out of 8 marbles in total. So, the probability of choosing a red marble on the second pick, given that the first pick was a red marble, is 2/8. To find the total probability of choosing two red marbles, we multiply the probabilities of each pick:
P(Method 1) = (3/9) * (2/8)
Method 2: Randomly choose a marble with replacement, then randomly choose another marble.
In this method, after choosing a marble, it is placed back in the bag, so the number of marbles remains the same for each pick. The probability of choosing a red marble on each individual pick is 3/9. Therefore, to find the total probability of choosing two red marbles, we multiply the probabilities of each pick:
P(Method 2) = (3/9) * (3/9)
By calculating both probabilities, we can determine which method has a greater probability of choosing two red marbles.
P(Method 1) = (3/9) * (2/8) = 6/72 = 1/12
P(Method 2) = (3/9) * (3/9) = 9/81 = 1/9
Comparing the two probabilities, we find that P(Method 2) has a greater probability of choosing two red marbles (1/9) compared to P(Method 1) (1/12). Therefore, the second method of randomly choosing a marble with replacement and then randomly choosing another marble has a greater probability of choosing two red marbles.