An urn contains 55 green marbles and 44 red marbles. One marble is removed, its color noted, and not replaced. A second marble is removed and its color is noted.
(a) What is the probability that both marbles are green? red?
(b) What is the probability that exactly one marble is green?
To solve this problem, we'll use the concept of probability.
(a) To find the probability that both marbles are green, we need to calculate the probability of drawing a green marble from the urn twice in a row.
The probability of drawing a green marble on the first pick is:
P(Green on first pick) = Number of green marbles / Total number of marbles
= 55 / (55 + 44)
= 55 / 99
Since we don't replace the first marble, the urn now contains 54 green marbles and 44 red marbles. So, the probability of drawing another green marble on the second pick is:
P(Green on second pick after a green) = Number of green marbles / Total number of marbles
= 54 / (54 + 44)
= 54 / 98
To find the probability of both marbles being green, we need to multiply the probabilities:
P(Both marbles are green) = P(Green on first pick) * P(Green on second pick after a green)
= (55 / 99) * (54 / 98)
≈ 0.301
Therefore, the probability that both marbles are green is approximately 0.301.
To find the probability that both marbles are red, we can follow a similar approach:
P(Red on first pick) = Number of red marbles / Total number of marbles
= 44 / (55 + 44)
= 44 / 99
Since we don't replace the first marble, the urn now contains 55 green marbles and 43 red marbles. So, the probability of drawing another red marble on the second pick is:
P(Red on second pick after a red) = Number of red marbles / Total number of marbles
= 43 / (55 + 43)
= 43 / 98
To find the probability of both marbles being red, we need to multiply the probabilities:
P(Both marbles are red) = P(Red on first pick) * P(Red on second pick after a red)
= (44 / 99) * (43 / 98)
≈ 0.197
Therefore, the probability that both marbles are red is approximately 0.197.
(b) To find the probability that exactly one marble is green, we need to consider two cases: either the first pick is green and the second pick is red, or the first pick is red and the second pick is green.
Case 1: First pick is green and second pick is red
P(Green-Red) = P(Green on first pick) * P(Red on second pick after a green)
= (55 / 99) * (44 / 98)
Case 2: First pick is red and second pick is green
P(Red-Green) = P(Red on first pick) * P(Green on second pick after a red)
= (44 / 99) * (55 / 98)
To find the probability that exactly one marble is green, we need to add the probabilities of the two cases:
P(Exactly one marble is green) = P(Green-Red) + P(Red-Green)
= (55 / 99) * (44 / 98) + (44 / 99) * (55 / 98)
≈ 0.564
Therefore, the probability that exactly one marble is green is approximately 0.564.
To find the probabilities, we need to use two concepts: probabilities and combinations.
(a) Probability that both marbles are green:
To find the probability of both marbles being green, we need to find the probability of the first marble being green and then multiply it by the probability of the second marble being green.
First, let's calculate the probability of drawing a green marble on the first draw:
There are a total of 99 marbles (55 green + 44 red) in the urn, so the probability of drawing a green marble on the first draw is 55/99.
After removing one green marble, there are 54 green marbles left out of a total of 98 remaining marbles (since we removed one). So, the probability of drawing another green marble on the second draw is 54/98.
To find the probability of both marbles being green, we multiply the two probabilities together:
P(Both marbles are green) = (55/99) * (54/98)
(b) Probability that exactly one marble is green:
To find the probability of exactly one marble being green, we need to consider two cases: either the first marble is green and the second is red, or the first marble is red and the second is green. We will add up the probabilities of these two cases.
Case 1: First marble is green, second is red:
The probability of drawing a green marble on the first draw is 55/99, as calculated before.
After removing a green marble, there are 44 red marbles left out of a total of 98 remaining marbles. So, the probability of drawing a red marble on the second draw is 44/98.
Case 2: First marble is red, second is green:
The probability of drawing a red marble on the first draw is 44/99 (opposite of the probability of the first case).
After removing a red marble, there are 55 green marbles left out of a total of 98 remaining marbles. So, the probability of drawing a green marble on the second draw is 55/98.
To find the probability of exactly one marble being green, we add the probabilities of the two cases together:
P(Exactly one marble is green) = (55/99) * (44/98) + (44/99) * (55/98)
Now you can calculate these probabilities using the given formulas.
99 marbles
first green = 55/99
second green = 54/98
so
(55/99)(54/98)
etc
for b
first green, second red plus first red, second green