A bag contains 5 red balls, 4 white balls, and 3 black balls. Two balls are drawn without
replacement. What is the probability that at least one ball drawn is white?
right. So what you want is 1 minus that.
So 1-.42= .58, so the probability of getting atleast 1 white ball is 58%?
well, what is the probability that no white ball is drawn?
I think 8/12 * 7/11= 56/132=42%?
To find the probability that at least one ball drawn is white, we can use the concept of complementary probability.
The total number of balls in the bag is 5 + 4 + 3 = 12.
Let's consider the case where no white ball is drawn. In this case, we have 7 non-white balls in the bag (5 red + 3 black), and we need to draw two balls without replacement.
To calculate the probability of drawing two non-white balls without replacement, we can use the formula for dependent events:
P(Event 1 and Event 2) = P(Event 1) * P(Event 2|Event 1)
In this case, Event 1 is drawing a non-white ball on the first draw, and Event 2 is drawing a non-white ball on the second draw, given that the first draw was a non-white ball.
P(Event 1) = (number of non-white balls) / (total number of balls) = 7 / 12
P(Event 2|Event 1) = (number of non-white balls after the first draw) / (total number of balls remaining) = 6 / 11
Using the formula for dependent events, we can calculate the probability of drawing two non-white balls without replacement:
P(no white ball) = P(Event 1 and Event 2) = P(Event 1) * P(Event 2|Event 1) = (7/12) * (6/11)
Now, the probability that at least one ball drawn is white is equal to 1 minus the probability of drawing no white ball:
P(at least one white ball) = 1 - P(no white ball)
P(at least one white ball) = 1 - [(7/12) * (6/11)]
Calculating this expression will give us the probability that at least one ball drawn is white.