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Given a bag of $4$ blue sweets and $2$ red sweets.

If I take out $4$ sweets at random, what's the probability of $2$ red sweets being a part of those $4$ picked out.

I'm pretty sure the answer is $0.4$, using a monte carlo simulation on my laptop, how would I get this mathematically?

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    Assume the sweets are numbered..$b_1,b_2,b_3,b_4,r_1,r_2$. There are $\binom 64$ equiprobable ways to choose $4$. There are $\binom 42$ ways to choose $4$ given that you must have $r_1,r_2$. – 2017-01-01

2 Answers 2

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First of all: I'll indicate with $\binom n k$ the number of possibilities of taking $n$ sweeties in groups of $k$ sweeties.

Now, we know that the total number of possibilities is $\binom 6 4 = 15$. The number of possibilities that 2 red sweeties are taken is $\binom 4 2 * \binom 2 2 = 6$.

So, the probability is $\frac 6 {15} = \frac 2 5 = 0.4$.

We can verify the result taking away, from the total number of possibilities, the number of possibilities that $0$ or $1$ red sweeties are taken, and they are $$\binom 4 4 = 1 $$ $$\binom 4 3 * \binom 2 1 = 4 * 2 = 8 $$

So we have $15 - (1+8) = 9$ possibilities, which gives us the probability $\frac 9 {15} = 0.6 \Rightarrow $ probability of taking 2 red sweeties = $1 - 0.6 = 0.4$.

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Probability that the $2$ sweets are red out of the $4$ sweets can be given by the ratio of

(number of ways of choosing $2$ blue sweets from $4$ after picking up the $2$ red sweets) $=\ ^4C_2$

and

(total number of choosing $4$ sweets from $6$) $=\ ^6C_4$.

Therefore: $$P=\frac{^4C_2}{^6C_4}=0.4$$