Question on probability/sets

Question

You have a basket containing 7 red balls, 5 green balls, and 2 orange balls. You pick 5 balls at random. What is the probability of getting at least one red ball?

Answer 1

Hi, here's one way that we can look at this:

You have a basket containing 7 red balls, 5 green balls, and 2 orange balls. You pick 5 balls at random. What is the probability of getting at least one red ball?

So, there are 14 balls in total. If you just pick one ball, the probability of getting a red ball is 7/14 (or 1/2), a green ball is 5/14, and an orange ball is 2/14 (or 1/7).

If we pick 5 balls at random, then the probability of at least one red ball can be looked at as the opposite of the probability of selecting "0" red balls, right?

So, the probability of selecting 0 red balls would be as follows:

1. Pick the first ball - the probability that it is not red is 7/14.
2. For the second ball, remember you now have 13 balls left (7 are still red, 6 not red). The probability, then, of picking a ball that is not red is 6/13. Let's continue this pattern.
3. Third ball, probability of not picking a red ball is 5/12
4. Fourth ball, probability of not picking a red ball is 4/11
5. Fifth ball, probability of not picking a red ball is 3/10

So, the probability of not picking a red ball in the first 5 picks is: (7/14)*(6/13)*(5/12)*(4/11)*(3/10) = 0.0105.,

Finally, the opposite of this is 1 - 0.0105 = 0.9895, or 98.95%, which is the probability of picking at least 1 red ball.

Answer 2

Find out the probability of getting no red balls, then subtract that probability from 1 to get the probability of getting at least one red ball

Use hypergeometric distribution to find probability of getting 0 red balls.

Will want to get zero red balls out of 7 available, and 5 balls out of the 7 remaining colors

will be sampling 5 out of the total 14 balls

C(7,0) * C(7,5) / C(14,5) = 0.01

P(getting at least 1 red ball) = 1 - 0.01 = 0.99

Answer 3

The answer depends on whether we replace the balls after we pick them. I think the wording subtly implies that we don't replace them, so I'll answer to that. It's often easier in these kind of probability questions involving the words "at least" to instead consider the probability you won't get a red ball. Then, the probability that you do get at least one will be the complement of that probability, 1-p.

In order to avoid getting a red ball, we have to hit a 7/14=1/2 chance to get a non-red ball, then a 6/13 chance as we've removed a non-red ball, and so on down to 3/10 for the 5th ball.

So, the probability of hitting these successive checks is (7/14)(6/13)(5/12)(4/11)(3/10)=3/286.

That's the probability of picking 5 NON-red balls in a row, which means that the probability of getting at least one red ball is 1-3/286= 283/286, or about 99%.

I also want to point out another way of answering the question, which may click better than the other.

We will still be pulling the trick of looking at the complement, but we'll be doing it from a different perspective.

The probability of getting no red balls will be equivalent to the number of 5-ball sets we can pull from the bag without red balls to the number of 5-ball sets we can pull out in general. To calculate the number of non-red sets we have, we can use the combination formula nCr, read "n choose r." In our case, we have "14 choose 5" sets of 5 we can make from 14 things. A calculator will tell you that this number turns out to be 2002.

So out of the 2002 possible 5-ball sets we could draw, how many have no red balls? Well, there are 7 non-red balls, so the number of no-red sets of 5 is "7 choose 5", or 21.

Thus, the probability is 21/2002=3/286. Finally, the probability of getting at least one red is 1-p(no red)=283/286 as we had before.

Answer 4

7+5+2 = 14 total balls. 7 out of 14 are red

use 14, 13, 12, 11 and 10 as the denominators and 7, 6, 5, 4 and 3 as the numerators of 5 fractions.

multiply them together,

then subtract that product from 1

probability of at least one red ball = one minus the probability of getting zero red balls

P(>0 Red) = 1 - P(0 Red) = 1 - (7/14)(6/13)(5/12)(4/11)(3/10)= 1- 2520/240240 = 1-.0105 = .9895

probability of at least one red in 5 draws = about 98.95%

Pr(>0 Red in 8 draws) = 1, as it's impossible not to get a red after the 1st 7 draws use up all the non-reds, on the 8th draw the only thing left are reds.

Get closer to 8 draws, such as 7, 6 or 5 draws , and you get closer to near certainty of at least one red, but slightly less than 100%

98.95% is the solution for the problem if the draws are without replacement.

If the problem was the probability if the draws are with replacement, the solution is 1-(1/2)^5 = 1- 1/32 = 96.825%

P(>0 in 8 draws) =1- (7/14)(6/13)(5/12)(4/11)(3/10)(2/9)(1/8)(0/7) = 1-0 = 1 = absolute certainty, even more certain than death & taxes, so to speak.

(although nothing is certain in this world, some things are if you assume things that really can never be absolutely assumed to be correct)