# Probability of 3rd ball being blue given at least 1 blue previously

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?

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

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

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

(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.

• And you can simplify : $P(E) = \frac{P(BBB)+2 \times P(BRB)}{1-P(RR)}$ Commented Jul 10 at 8:07
• 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
– Prem
Commented Jul 10 at 10:25