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.
Best Answer
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 $$ P(R2)=P(R1)*P(R2|R1)+P(G1)*P(R2|G1) $$ $$ =6/16*5/15 + 10/16*6/15 $$ $$ =3/8 $$