A bag contains 5 red and 4 green discs , three discs are randomly drawn from a bag what is the probability at least 1 is red and the probability at most one is green?
find probability that all are green
first one 4/9
second one 3/8
third one 2/7
all green = (4/9)(3/8)(2/7)
answer = 1 - that result
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find probability that 9 of 9 are red
find probability that 8 of 9 are red
that sum
is the probability that zero or one is green
To find the probability that at least one disc is red when three discs are randomly drawn from the bag, we will calculate the probability of the complementary event and subtract it from 1.
Step 1: Calculate the probability that all three discs are green.
The probability of drawing a green disc on the first draw is 4/9.
After the first green disc is drawn, there are 8 discs left in the bag, with 3 of them being green.
So, the probability of drawing a green disc on the second draw given that the first draw was green is 3/8.
Similarly, the probability of drawing a green disc on the third draw given that the first two draws were green is 2/7.
Therefore, the probability of drawing three green discs in a row is (4/9) * (3/8) * (2/7).
Step 2: Subtract the probability of all three discs being green from 1 to find the probability of at least one disc being red.
P(at least one red) = 1 - P(all green)
P(at least one red) = 1 - (4/9) * (3/8) * (2/7)
Now, to find the probability of at most one green disc, we need to consider the cases where there are zero green discs and one green disc.
Case 1: Zero green discs (all red)
The probability of drawing a red disc on the first draw is 5/9.
After the first red disc is drawn, there are 4 red discs left in the bag, out of a total of 8 discs.
So, the probability of drawing a red disc on the second draw given that the first draw was red is 4/8.
Similarly, the probability of drawing a red disc on the third draw given that the first two draws were red is 3/7.
Therefore, the probability of drawing three red discs in a row is (5/9) * (4/8) * (3/7).
Case 2: One green disc and two red discs
The probability of drawing a green disc on the first draw is 4/9.
After the first green disc is drawn, there are 8 discs left in the bag, with 4 of them being red.
So, the probability of drawing a red disc on the second draw given that the first draw was green is 4/8.
Similarly, the probability of drawing a red disc on the third draw given that the first two draws were green is 3/7.
Therefore, the probability of drawing one green disc and two red discs is (4/9) * (4/8) * (3/7).
Step 3: Add the probabilities of the two cases to find the probability of at most one green disc.
P(at most one green) = P(all red) + P(one green, two red)
P(at most one green) = (5/9) * (4/8) * (3/7) + (4/9) * (4/8) * (3/7)
Finally, the probability that at least one disc is red is 1 - P(all green), and the probability of at most one green disc is P(all red) + P(one green, two red).
To find the probability, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
Let's start by finding the probability that at least one disc is red.
To do this, we need to calculate the number of favorable outcomes (at least one red disc) and divide it by the total number of possible outcomes (drawing three discs from the bag without any restriction).
The number of favorable outcomes can be calculated by considering all possible scenarios:
1. The first disc is red, and the remaining two discs can be any color: 5 * (9 choose 2) = 5 * 36 = 180
2. The second disc is red, and the remaining two discs can be any color: 4 * 5 * (8 choose 1) = 4 * 5 * 8 = 160
3. The third disc is red, and the remaining two discs can be any color: 4 * 5 * (8 choose 2) = 4 * 5 * 28 = 560
Now, let's calculate the total number of possible outcomes (drawing three discs from the bag without any restriction):
9 choose 3 = (9!)/((3!)(9-3)!) = (9!)/(3!6!) = 84
Therefore, the number of favorable outcomes is 180 + 160 + 560 = 900, and the total number of possible outcomes is 84.
Therefore, the probability of drawing at least one red disc is 900/84 ≈ 0.7143.
Now let's calculate the probability that at most one disc is green.
To do this, we need to calculate the number of favorable outcomes (at most one green disc) and divide it by the total number of possible outcomes (drawing three discs from the bag without any restriction).
The number of favorable outcomes can be calculated by considering the following scenarios:
1. No green discs are drawn: This scenario has only one possibility, which is drawing three red discs. The number of ways to choose three red discs from five is 5 choose 3 = (5!)/((3!)(5-3)!) = 10.
2. One green disc is drawn: We need to calculate the number of ways to choose one green disc and two red discs. The number of ways to do this is (4 choose 1) * (5 choose 2) = 4 * 10 = 40.
Now, let's calculate the total number of possible outcomes (drawing three discs from the bag without any restriction), which is 84 (as calculated earlier).
Therefore, the number of favorable outcomes is 10 + 40 = 50, and the total number of possible outcomes is 84.
Therefore, the probability of drawing at most one green disc is 50/84 ≈ 0.5952.
So, the probability of drawing at least one red disc is approximately 0.7143, and the probability of drawing at most one green disc is approximately 0.5952.