You have a bucket with 3 black balls and 7 red balls. What is the probability of drawing 2 reds and 1 black in 3 draws if order is not important? (Assume replacement after each draw)
ways to do this:
RRB
RBR
BRR
so three ways, then Pr= 3*7/10*7/10*3/10
So Bob - Final answer would be .441?
To find the probability of drawing 2 red balls and 1 black ball in 3 draws without considering the order, we need to calculate the probability of each possible combination of red and black balls.
First, let's calculate the probability of drawing a red ball. There are 7 red balls out of a total of 10 balls in the bucket, so the probability of drawing a red ball on the first draw is 7/10.
Since we're replacing the balls after each draw, the probability of drawing a second red ball is also 7/10.
Now, let's calculate the probability of drawing a black ball. There are 3 black balls out of a total of 10 balls in the bucket, so the probability of drawing a black ball on the first draw is 3/10.
Since we're replacing the balls after each draw, the probability of drawing a second black ball is also 3/10.
Now, we need to consider the order in which the balls can be drawn. Since we're not considering order, we need to account for the different combinations of red and black balls.
There are three possible combinations: two reds and one black, one red and two blacks, and three blacks. Since we're interested in the probability of drawing two reds and one black, we only need to calculate the probability of the first two draws being red and the third draw being black.
The probability of drawing two reds and one black is calculated by multiplying the probabilities of each event happening:
(7/10) * (7/10) * (3/10) = 147/1000 = 0.147
Therefore, the probability of drawing 2 reds and 1 black in 3 draws (without considering the order) is 0.147, or 147/1000.