if i have 2 red candies 1 green and 1 purple and i’m picking two at a time, what’s the possibility of getting two reds?
Explain please.
There are four equally-likely possible outcomes:
(a) 2 red are picked
(b) a red and a green are picked
© a red and a purple are picked
(d) a green and a purple are picked
Therefore, 25%.
It’s a roll of the die - 1 in 6.
You can always just count the possibilities (especially easy when there are so few options).
Even though there are 2 red candies, you should count each separately when enumerating. There’s a problem with saying 1 in 4. There are 4 outcomes but they are NOT equally likely:
(A) Two red - exactly one way to do this
(B) 1 red, 1 green - two ways to do this (each red) - it’s twice as likely as (A).
(C) 1 red, 1 purple - two ways to do this (each red) - it’s twice as likely as (A).
(D) 1 green, 1 purple - exactly one way to do this.
So, of the 4 outcomes, only 1 (A) matches what you want, but it is half as likely as (B) and (C). For purposes of computing a probability, it’s easier to list each red separately.
So, we have Red A, Red B, Green, and Purple.
The ways to pick 2 of these are:
Red A, Red B (what we want)
Red A, Green
Red B, Green
Red A, Purple
Red B, Purple
Green, Purple
So, there are 6 ways of choosing 2 candies and only 1 of them match our criteria.
For a more formal way of expressing exactly the same information, there are C(4,2) combinations of 4 things taken 2 at a time, or:
C(4,2) = 4! / (2! * (4-2)!)
= 4 ! / (2!2!)
= 24 / (22)
= 24/4
= 6
So, there are 6 total ways of picking 2 of the candies.
Of these, there are C(2,2) ways of picking 2 out of 2 red candies. C(2,2) = 1, so that’s simple.
There are also C(2,0) ways of picking 0 out of the 2 non-red candies C(2,0) = 1, so that’s also simple.
So, there are C(2,2)*C(2,0) ways of picking 2 of the 2 red candies and 0 of the 2 non-red candies. In this case, that’s 1 way, as C(2,2)C(2,0) = 11 = 1.
So, the overall probability is 1 / 6, as we noted earlier.
The same concept can be used to compute many similar sorts of probabilities, including lotto probabilities.
Six equally-likely picks (r1 r2, r1 g, r1 p, r2 g, r2 p, g p), so 1/6 chance.
[ninjaed]
In short, to list the 4 possible outcomes with probabilities, we have:
(A) 2 red - 1/6
(B) 1 red, 1 green - 2/6 = 1/3
© 1 red, 1 purple - 2/6 = 1/3
(D) 1 green, 1 purple - 1/6
Sorry to keep posting, but there’s another way of expressing the idea these are not equally likely outcomes.
Say we have 4 people from 3 different states:
Tom and Anne from Texas
Bill from Montana
Susan from Kentucky
If randomly chosen, what is the probability we pick the two from Texas?
Tom and Bill is a different from Anne and Bill, even though both are “Texas + Montana”, so each should be counted separately for probability purposes. The Red candies are no different.
Very helpful. thanks,
Just want to point out that it doesn’t matter if you are choosing them one at a time or together: the odds of the final outcome would be the same.
Right, and this gives us an alternative way of approaching the problem:
The probability that the first candy selected is a red one is 2/4 (= 1/2).
The probability that the second candy is red if the first one was is 1/3 (since we’re selecting from among the three remaining).
So, by the multiplication rule for dependent events, the probability that the first and second are both red is
(1/2)x(1/3) = 1/6.
The above illustrates how easy it is to have what looks like a really good argument about probability and be totally wrong.
That’s how I attacked it, getting 1/6 in just a moment’s thought. But I wasn’t sure I was right until reading the replies, even for such a simple problem. Probability stuff is SO easy to get wrong!
This is a good example of a hypergeometric distribution, the one you use to compute odds of winning the Lotto.
Ignore the colors for a moment and just label the four candies A, B, C, and D. There are six ways you can pick two of them: AB, AC, AD, BC, BD, CD. If two of them are red, say A and B, then only 1/6 of the possibilities give you both reds.
Ahh, it is 1 in 6, not 1 in 4. I stand corrected, thanks guys.
100%. Or 0% if you don’t feel like picking two reds I guess. I don’t know, maybe we should condition the problem based on some prior about the chooser’s color preferences?
… what? The question never specified I was picking them at random with no idea of which had which color. 
We beat that one to death.
1/2 x 1/3 is 1/6
Picking two reds is a certainty if you’re allowed to look. Which level of Candy Crush Saga did you find this puzzle?