Daniel will randomly pick a marble from a bag that has three red marbles and six white. If he does not put any marble back into the bag and picks a new one each time, what is the probability that both marbles will be white. If he puts the first marble back into the bag and picks another, what is the probability that both are white?
Prob(2 white without replacement)
= (6/9)(5/8)
= 5/12
prob(2 white with replacement)
= (6/9)(6/9)
= 4/9
Ah, the old marble dilemma! Let's roll with it, shall we?
If Daniel doesn't put the marble back in the bag after each pick, the probability of getting a white marble on the first pick is 6 out of 9. Now, since he doesn't replace it, he has one less marble to choose from (5 white marbles left out of 8 total marbles). Thus, the probability of getting a white marble on the second pick is 5 out of 8. To find the probability of both marbles being white, we need to multiply these probabilities. So, the probability is (6/9) * (5/8) = 30/72 = 5/12.
Now, if Daniel places the first marble back into the bag, he has the same probability of picking a white marble on the first pick, which is 6 out of 9. Since he puts it back, the bag remains unchanged with 6 white marbles out of 9 total. Therefore, the probability of getting a white marble on the second pick is again 6 out of 9. To find the probability of both marbles being white, we multiply these probabilities, resulting in (6/9) * (6/9) = 36/81 = 4/9.
So, to recap: the probability of both marbles being white, without replacement, is 5/12, and with replacement, it is 4/9.
To calculate the probability, we need to consider two scenarios: one where Daniel does not put the first marble back into the bag, and another where he does put it back.
Scenario 1: Daniel does not put the first marble back into the bag
In this case, after Daniel picks the first marble, there will be two white marbles remaining out of a total of eight marbles (since he is not replacing the first marble).
The probability of picking a white marble on the second pick is 2/8 or 1/4, since there are now two white marbles out of the remaining eight.
Therefore, the probability that both marbles will be white is (6/9) * (2/8) = 1/6.
Scenario 2: Daniel puts the first marble back into the bag
In this case, after Daniel picks the first marble, there will still be six white marbles remaining out of a total of nine marbles (since he replaces the first marble).
The probability of picking a white marble on the second pick is still 6/9, since the composition of the bag remains the same.
Therefore, the probability that both marbles will be white is (6/9) * (6/9) = 4/9.
To summarize:
- If Daniel does not put the first marble back, the probability that both marbles will be white is 1/6.
- If Daniel puts the first marble back, the probability that both marbles will be white is 4/9.
To determine the probability in each scenario, we need to calculate the probability of drawing a white marble for each selection.
1. Without putting the marble back into the bag:
Since there are a total of 9 marbles (3 red and 6 white), the probability of drawing a white marble on the first pick is 6/9.
After the first marble is drawn, there are now 8 marbles left in the bag (2 red and 6 white), so the probability of drawing a white marble on the second pick is 6/8.
To find the probability of both marbles being white, you need to multiply the probabilities of each independent event. Therefore, we multiply (6/9) * (6/8) = 36/72 simplify to 1/2.
So, without putting the first marble back, the probability of drawing two white marbles is 1/2 or 0.5.
2. By putting the marble back into the bag:
The probability of drawing a white marble on the first pick is still 6/9.
But after the first marble is drawn, it is put back into the bag, so we have the same initial probability of 6/9 for the second pick.
Again, we multiply the probabilities of each individual event: (6/9) * (6/9) = 36/81 simplify to 4/9.
Therefore, if he puts the first marble back into the bag, the probability of drawing two white marbles is 4/9 or approximately 0.44.