Calculate the probability of obtain a red ball from a urn without replacement

Question

We have a urn with six red balls and ten green balls. Two balls are randomly selected in orden from the urn without remplacement. Calculate the probability that the second ball is red.

Because obtain a red ball (R) is a event independent from obtain a green ball (G), we have that $$P(R_{2}) = P(R_{2}|R_{1}) + P(R_{2}|G_{1})$$ Then, we have that $P(R_{2}|R_{1})= \frac{5}{15}$ because we have the first ball is red, so we had in the urn $15$ balls in total and $5$ are red. Similarly, we have that $P(R_{2}|G_{1}) = \frac{6}{15}$.

Therefore $$P(R_{2})= \frac{11}{15}$$

Am I wrong? I don't think I got it right.

Answer 1

I think your answer is wrong because you forgot to multiply the probability of the what you get as the first ball. So The Equation Should Be $$ P(R2)=P(R1)*P(R2|R1)+P(G1)*P(R2|G1) $$ $$ =6/16*5/15 + 10/16*6/15 $$ $$ =3/8 $$

Answer 2

You have the following possible cases

$$RR - RG - GR - GG$$

Your probability is

$$P(RR)+P(GR)=\frac{6}{16}\cdot\frac{5}{15}+\frac{10}{16}\cdot\frac{6}{15}=\frac{3}{8}$$

Answer 3

There are $10$ green (G) and $6$ red (R) balls and there are two ways you can get second ball as red - GR or RR

Probability of GR is $\frac{10}{16} \cdot \frac{6}{15} = \frac{1}{4}$

Probability of RR is $\frac{6}{16} \cdot \frac{5}{15} = \frac{1}{8}$

So the probability that the second ball is red is,

$\displaystyle \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$

Answer 4

Why don’t we assign the second ball first (6 red balls possible) and then the first ball (15 balls remaining). Then the probability is simply:

$$ \frac{6\times 15}{16\times 15}=\frac{3}{8} $$