In a bag there are 2 red marbles 3 white marbles and 5 blue marbles. Once a marbel is picked it is not replaced. Find the following probabilities:
P(red,then white)
P(blue, then red)
P(red, red, red)
P(blue, blue, white)
But how would you solve the other 3 then
I still don't understand.
That makes 0 sense there’s 10 balls not 12 just say the answer
To find the probabilities, we need to use the concept of conditional probability, where the probability of an event happening is dependent on the outcome of a prior event. We will apply this concept to solve the different scenarios you provided.
1. P(red, then white):
To find the probability of picking a red marble first and then a white marble, we need to consider the fact that once a marble is picked, it is not replaced. Since the marbles are not replaced, the total number of marbles decreases each time we pick one.
The probability of picking a red marble first is 2/10 (as there are 2 red marbles out of a total of 10 marbles initially).
Then, the probability of picking a white marble given that we have already picked a red marble is 3/9 (as there are 3 white marbles remaining out of a total of 9 marbles).
Therefore, the probability of picking a red marble first and then a white marble is (2/10) * (3/9) = 6/90 = 1/15.
2. P(blue, then red):
Similar to the previous scenario, we consider that the marbles are not replaced after being picked.
The probability of picking a blue marble first is 5/10 (as there are 5 blue marbles out of a total of 10 initially).
Then, the probability of picking a red marble given that we have already picked a blue marble is 2/9 (as there are 2 red marbles remaining out of a total of 9 marbles).
Therefore, the probability of picking a blue marble first and then a red marble is (5/10) * (2/9) = 10/90 = 1/9.
3. P(red, red, red):
To find the probability of picking three red marbles in a row, we again consider that the marbles are not replaced.
The probability of picking the first red marble is 2/10.
Given that we have already picked a red marble, the probability of picking another red marble is 1/9 (as there would be only 1 red marble remaining out of a total of 9 marbles).
Finally, the probability of picking the third red marble, given that we have already picked two red marbles, is 0/8 (as no red marbles would be left to pick).
Therefore, the probability of picking three red marbles in a row is (2/10) * (1/9) * (0/8) = 0.
4. P(blue, blue, white):
To find the probability of picking a blue marble, followed by another blue marble, and then a white marble, we apply the same logic.
The probability of picking the first blue marble is 5/10.
Given that we have already picked a blue marble, the probability of picking another blue marble is 4/9 (as there would be 4 blue marbles remaining out of a total of 9 marbles).
Finally, the probability of picking a white marble, given that we have already picked two blue marbles, is 3/8 (as there are 3 white marbles remaining out of a total of 8 marbles).
Therefore, the probability of picking a blue marble, then another blue marble, and finally a white marble is (5/10) * (4/9) * (3/8) = 60/720 = 1/12.
So, the probabilities are:
P(red, then white) = 1/15
P(blue, then red) = 1/9
P(red, red, red) = 0
P(blue, blue, white) = 1/12.
all are basically the same. The total number of marbles drops after each selection. For example,
P(blue, blue,white) = 5/10 * 4/9 * 3/8
for each draw, it is the number of that color, divided by the total number of marbles. For more than one draw, just multiply the fractions.
Take a look at what I did. It should become clear.
Look. If there are 12 balls, and 5 are red, the chance of getting a red ball is 5/12.
After you have drawn that red ball, there are only 4 reds and 11 balls, so the chance of getting a 2nd red is 4/11
and so on.