A gumball machine contains 6 red gumballs and 6 white gumballs.
Two gumballs are purchased, one after the other, without replacement.
Find the probability that at least one gumball is white.
Express your answer as a decimal, rounded to the nearest hundredth.
To find the probability that at least one gumball is white, we can calculate the probability that both gumballs are red and subtract it from 1.
Step 1: Calculate the probability of the first gumball being red.
The probability of the first gumball being red is 6 red gumballs out of a total of 12 gumballs, so the probability is 6/12 or 1/2.
Step 2: Calculate the probability of the second gumball being red.
After one gumball is already removed, there are only 11 gumballs left in the machine, with 5 of them being red. Therefore, the probability of the second gumball being red is 5/11.
Step 3: Calculate the probability that both gumballs are red.
To find this probability, we multiply the probabilities from steps 1 and 2 together:
Probability = (1/2) * (5/11) = 5/22.
Step 4: Calculate the probability that at least one gumball is white.
Since we want the complement of both gumballs being red, we subtract the probability calculated in step 3 from 1:
Probability = 1 - 5/22 = 17/22.
Therefore, the probability that at least one gumball is white is 17/22, which rounded to the nearest hundredth is 0.77.
To find the probability that at least one gumball is white, we can calculate the probability of the complementary event and subtract it from 1. The complementary event in this case is the event where both gumballs drawn are red.
To calculate the probability of drawing two red gumballs, we need to consider the number of red and white gumballs remaining after each draw.
For the first draw, there are 6 red and 6 white gumballs, so the probability of drawing a red gumball is given by:
P(Drawing a red gumball on the first draw) = 6 red gumballs / (6 red gumballs + 6 white gumballs) = 6/12 = 1/2
After the first draw, there will be 5 red gumballs and 6 white gumballs remaining in the machine. So, for the second draw, the probability of drawing a red gumball is given by:
P(Drawing a red gumball on the second draw) = 5 red gumballs / (5 red gumballs + 6 white gumballs) = 5/11
To find the probability of both draws resulting in red gumballs, we multiply the individual probabilities together:
P(Both draws resulting in red gumballs) = P(Drawing a red gumball on the first draw) * P(Drawing a red gumball on the second draw) = (1/2) * (5/11) = 5/22 ≈ 0.2273
Now we can find the probability of at least one gumball being white by subtracting the probability of both draws resulting in red gumballs from 1:
P(At least one gumball is white) = 1 - P(Both draws resulting in red gumballs) = 1 - 5/22 = 17/22 ≈ 0.7727
Therefore, the probability that at least one gumball is white is approximately 0.77 when rounded to the nearest hundredth.
well, what is the probability that NO ball is white? That is, 2 reds?
Then take 1-P(red,red)