An urn contains six red balls and five white balls. A sample of 2 balls is selected at random from the urn. What is the probability that at least one of the balls is red?
This is the probability of one red ball or two red balls.
One = 6/11 * 5/10 = ?
Two = 6/11 * 5/10 = ?
Either-or probability is found by adding the individual probabilities.
Well, statistically speaking, there are only two types of people in this world - those who love probability and those who just don't get it. But don't worry, I'm here to make it all clear for you!
To find the probability that at least one ball is red, we need to consider the different scenarios. There are a couple of ways this can happen: either we pick one red ball and one white ball, or we pick two red balls.
Let's break it down step by step:
Scenario 1: Picking one red ball and one white ball
The probability of picking a red ball first is 6/11 (since there are 6 red balls out of 11 total).
Once we've picked a red ball, we're left with 5 red balls and 5 white balls in the urn.
The probability of picking a white ball next is 5/10 (since there are 5 white balls left out of 10 total).
So the probability of this scenario happening is (6/11) * (5/10).
Scenario 2: Picking two red balls
The probability of picking a red ball first is still 6/11.
And once we've picked a red ball, the probability of picking another red ball is 5/10.
So the probability of this scenario happening is (6/11) * (5/10).
Now, all that's left to do is add up the probabilities of these two scenarios to get the total probability that at least one ball is red.
So the final answer is (6/11) * (5/10) + (6/11) * (5/10) = (6/11) * (1/2) + (6/11) * (1/2) = (6/11) * (1) = 6/11.
So, the probability that at least one of the balls is red is 6/11.
To find the probability that at least one of the balls is red, we can find the probability of the complementary event, which is the probability that both balls are white, and subtract it from 1.
The total number of ways to select 2 balls from 11 balls (6 red and 5 white) is given by the binomial coefficient:
C(11, 2) = 11! / (2! * (11-2)!) = 55
The number of ways to select 2 white balls from the 5 white balls is:
C(5, 2) = 5! / (2! * (5-2)!) = 10
Therefore, the probability of selecting 2 white balls is:
P(2 white) = 10/55
The probability of the complementary event, where at least one ball is red, is:
P(at least one red) = 1 - P(2 white)
P(at least one red) = 1 - (10/55)
Simplifying, we get:
P(at least one red) = 45/55
So, the probability that at least one of the balls is red is 45/55, which can be further simplified to:
P(at least one red) = 9/11.
To find the probability that at least one of the balls selected is red, we need to calculate the probability of the complement event and subtract it from 1.
Step 1: Find the probability of selecting two white balls.
The probability of selecting a white ball on the first draw is 5 out of 11, since there are 5 white balls and 11 total balls in the urn. After one white ball has been selected, there are 4 white balls left out of 10 total balls. So, the probability of selecting another white ball is 4 out of 10. We need to multiply these probabilities together since the events are independent.
P(white on the first draw) = 5/11
P(white on the second draw) = 4/10
P(white on both draws) = P(white on the first draw) * P(white on the second draw) = (5/11) * (4/10) = 20/110 = 2/11
Step 2: Calculate the probability of selecting at least one red ball.
The complement event of selecting at least one red ball is selecting two white balls, which we found in Step 1. Therefore, the probability of selecting at least one red ball is 1 minus the probability of selecting two white balls.
P(at least one red ball) = 1 - P(white on both draws)
P(at least one red ball) = 1 - (2/11) = (11/11) - (2/11) = 9/11
So, the probability that at least one of the balls selected is red is 9/11 or approximately 0.8182.