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I have a bin with $5$ blue and $10$ red balls. I take out $2$ and at least one is blue. I then take a 3rd.
What is the probability that the 3rd ball is blue?

I understand I likely need conditional probability but I'm having a hard time with the "at least 1 blue" because whether it's 1 or 2 affects the 3rd probability.

Is this the same question as if I took out 3 and then looked at them 1 by 1?

Can this be done just as follows?

$$P(BBB) + P(BRB) + P(RBB) = \frac{5}{15}\times\frac{4}{14}\times\frac{3}{13} + \frac{5}{15}\times\frac{10}{14}\times\frac{4}{13} + \frac{10}{15}\times\frac{5}{14}\times\frac{4}{13}$$

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  • $\begingroup$ Use $P(X|Y)P(Y)=P(X,Y)$. $\endgroup$ Commented Jul 10 at 1:50

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You are on the right track , though that is not the Probability we want.

What we want is Conditional Probability like this :

$$P(E) = \frac{P(BBB) + P(BRB) + P(RBB)}{P(BBB) + P(BRB) + P(RBB) + P(BBR) + P(BRR) + P(RBR)}$$

[[ $E$ is the "Event" concerned : third Ball Blue , given at least 1 Blue Ball among first & second ]]

We can then make it :
$$P(E) = \frac{P(BBB) + P(BRB) + P(RBB)}{P(BBX) + P(BRX) + P(RBX)}$$

[[ $X$ indicates either ]]

Hence it is just :
$$P(E) = \frac{P(BBB) + P(BRB) + P(RBB)}{P(BB) + P(BR) + P(RB)}$$

ADDENDUM :

(1) Probability OP gave will match the Event "take 3 Balls , at least 3 Balls are Blue , last Ball is Blue" , which is not Conditional Probability

(2) When we consider all Events , Probability should add up to Exactly $1.0$ always , where the Solution given here will work out : Add that Probability to the Event that the last Ball is Red & we will indeed get $1.0$ Consistently.
The value given by OP will add up to very tiny total , which will alert us that something is wrong.

(3) When Events (like taking 2 Balls out) have already occurred , the Sample Space will change accordingly , incorporating those states.
We have to work within the new Sample Space.
That is Conditional Probability in a nut-shell.

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  • $\begingroup$ And you can simplify : $P(E) = \frac{P(BBB)+2 \times P(BRB)}{1-P(RR)}$ $\endgroup$
    – Lourrran
    Commented Jul 10 at 8:07
  • $\begingroup$ That is Correct here : Denominator Simplification is almost universally valid , @Lourrran , though Numerator Simplification is valid only when we have $BBR = BRB = RBB$ , like with the OP Experiment $\endgroup$
    – Prem
    Commented Jul 10 at 10:25

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