0

I'm attempting the same problem as posted here:

A jar contains $r$ red balls and $g$ green balls, where $r$ and $g$ are fixed positive integers. A ball is drawn from the jar randomly (with all possibilities equally likely), and then a second ball is drawn randomly.

Suppose that there are 16 balls in total, and that the probability that the two balls are the same colour is the same as the probability that they are different colours. What are $r$ and $g$ (list all possibilities)?

The claim is that the probability that the two balls are different colors is $1/2$, which means that the probability that the two balls are the same color is also $1/2$.

However, it seems to me that the probability would be $1/4$, since the probability of getting any color in both scenarios is $1/2$, which means that $1/2 \cdot 1/2 = 1/4$?

I would greatly appreciate it if people could please take the time to explain why I'm wrong and why the author is correct.

Community
  • 1
The Pointer
  • 4,182

1 Answers1

1

We generally have two distinct cases:$$A=\text{the balls have different colors}\\B=\text{the balls have same color}$$therefore$$A\cap B=\emptyset\\A\cup B=U$$where $U$ denotes the Universal Set. The probability that we draw out two balls with the same color does not remain $1\over 2$ in both draws as we do this without replacement. Also there are exactly two cases for each scenario ($A$ or $B$): for $A$ we have $(r,r)$ and $g,g$ and for $B$, $(r,g)$ and $(g,r)$

Mostafa Ayaz
  • 31,924
  • Thanks for the answer. Sorry if I'm misinterpreting your answer, but it seems like you're saying that the author is incorrect to claim that the probability is $1/2$? Please clarify. – The Pointer Aug 18 '19 at 13:20
  • 1
    Call the probability to draw red to be R and green to be G. Then, drawing the same twice is R^2 + G^2 and drawing different is 2RG where R+G = 1. So R^2 + G^2 = 2RG, and since their sum must be one (they are mutually exclusive and represent the set of all probabilities), they must both be half. In other words, the author is right in that the chances of drawing same vs different are both half. – Gabe Aug 18 '19 at 13:27
  • 2
    You're welcome. The author is right for what he said. Based on his claim $$P(A)=P(B)$$and since $$P(A)+P(B)=1$$then $$P(A)=P(B)={1\over 2}$$ – Mostafa Ayaz Aug 18 '19 at 13:29
  • @MostafaAyaz Ahh, yes, it's clear now. Thank you again! – The Pointer Aug 18 '19 at 13:37