A bag contains 5 green marbles,8 red marbles,11 orange marbles, 7 brown marbles, and 12 blue marbles. You choose a marble, replace it, and choose again. What is P(red, then blue)?
A.20/43
B.40/43
C.20/1849
D.96/1849
Is the answer A?? Thank you in advance
Total marbles = 43
P(red) = 8 / 43
Since it was replaced, the total number of marbles is the same for the second choosing of marble.
P(blue) = 12 / 43
Probability of getting a red, then a blue:
8/43 * 12/43 = 96/1849 (letter D)
Thank you
I would think it was A because you replaced the red one and still had 43 marbles. You would just add the 8+12 which would be 20/43.
Well it's a conditional probability. You have to multiply the probability of the first event to the probability of the second event for this kind of problem.
If the question asked for the probability of choosing a red OR a blue marble, then their probabilities are added.
But the question asked for the probability of getting a red marble (first choosing), and then choose again to get blue marble (second choosing).
If it's still not clear and you want some examples, then I suggest you read lectures about 'conditional probability'.
hope this helps~ `u`
To calculate the probability of selecting a red marble, replacing it, and then selecting a blue marble, we need to multiply the probability of both events occurring.
First, let's calculate the probability of selecting a red marble. There are a total of (5 + 8 + 11 + 7 + 12) = 43 marbles in the bag, and 8 of them are red. So, the probability of selecting a red marble on the first draw is 8/43.
Since the marble is replaced after each draw, the probability of selecting a blue marble on the second draw is also 12/43.
To find the probability of both events occurring, we multiply their individual probabilities:
P(red, then blue) = (8/43) * (12/43)
Simplifying this expression, we get:
P(red, then blue) = 96/1849
So, the correct answer is D. 96/1849, not A. 20/43.