# Probability of drawing at least 2 blue balls

There are 3 types of balls: Red, green and blue. Each of them has equal probability of being drawn. What is the probability of having at least two blue balls out of 5 balls. Can somebody correct me since this should not be right answer(answer is $$P(A) = \frac{10}{21} \approx0.476$$).

The total number of possible outcomes |$$\Omega$$| = $$3^{5}$$

Let A be event of drawing at least 2 blue balls, therefore $$A^c$$ - drawing 0 i or 1 blue ball

for 0 blue balls: $$2^5$$

for 1 blue ball: $$5\choose1\times2^4$$

$$P(A^c) = \dfrac{2^5 + {5\choose1}\times2^4}{3^5} \approx 0.46$$

$$P(A) = 1 - 0.46 = 0.54$$

• You can consider using binomial probability for this question. Commented Nov 23, 2019 at 11:44
• Anyway, I think your solution is correct. Your textbook may have an error. Commented Nov 23, 2019 at 11:53
• Your problem seems strangely posed. Do you put the ball back when you draw it or not ? Equivalently, Do you have an infinite amount of balls in that bag, so the probability of drawing a blue ball stays the same ? Or is the probability of drawing a blue ball affected by drawing a blue ball, or by drawing other ball colors ? Commented Nov 23, 2019 at 12:13
• Ye i know its weird, its like a shop sells three types of balls in a box of 5 (random balls with equal probability). What is the probability you bought a box with at least 2 blue balls? Commented Nov 23, 2019 at 12:19
• I can't find any error in your approach. The answer key must be wrong.
– Sam
Commented Nov 23, 2019 at 13:33

## 1 Answer

The answer $$\frac{10}{21}$$ is obtained in assumption that all combinations of $$x_1$$ red balls, $$x_2$$ green balls and $$x_3$$ blue balls with $$x_1+x_2+x_3=5$$, $$x_i\geq 0$$, $$i=1,2,3$$ are equally probable. That is, we can get $$5$$ red balls, or $$4$$ red and $$1$$ blue, or $$2$$ red, $$2$$ blue and $$1$$ green and so on, and all these cases have the same probability.

This is the scheme where $$5$$ indistinguishable balls are placed into $$3$$ distinct boxes R,G,B. It is not clear why this scheme is used for the given answer and how it can follow from the phrase

Each of them has equal probability of being drawn.

I think that this problem is poorly worded. And the solution of OP is right.

Newertheless, if we assume that we place $$5$$ indistinguishable balls into $$3$$ distinct boxes R,G,B, then total number of outcomes is $$|\Omega|=\binom{5+3-1}{3-1}=\binom{7}{2}=21$$ and a number of outcomes with at least two blue balls is $$|A|=\binom{(5-2)+3-1}{3-1}\binom{5}{2}=10.$$

• But it would be a very weird distribution to have the distinguishable stars-and-bars outcomes equally likely -- that never happens in ordinary models of randomly placing objects in boxes. That's a serious conceptual error in the book from which this problem is drawn ... nice work finding it!
– Ned
Commented Nov 23, 2019 at 19:24