0
$\begingroup$

A few weeks ago, I and my friend came across a question in some old maths book of him which is as follows.

A box contains 6 new balls and 4 old balls. 2 balls are drawn out from the box at a time. 1 of them is found to be a new ball. What is the probability that the other one is also a new ball?

Both of us tried to answer it. When we completed we found two different answers.

  1. First answer (written by my friend)

    • Probability of 1 new ball drawn, P(A) = 6C1/10C1 = 6/10.

    • Probability of drawing both the balls as newer ones, P(A intersection B) = 6C2/10C2 = 1/3.

    • Probability of drawing the second new ball provided with the first ball new, P(B/A) = P(A intersection B) / P(A) = 1/3 * 10/6 = 5/9.

  2. Second answer (written by me)

    • Out of the two balls to be drawn the first one is new. So, we have two exhaustive cases now. One is that the second ball must be old. The other one is that it must be new.
    • Total number of exhaustive events = 2.

    • Total number of favourable cases ( chances of the second ball to be good) = 1.

    • Probability = Total number of favourable cases / Total number of exhaustive cases = 1/2.

I say that the answer of mine is right. And he says that his answer is right. Both of them don't know to explain the other why the stand of ours is correct. Can somebody tell us which of the answers is right? Do give us an explanation that can make us understand the concept well enough.

$\endgroup$
1
$\begingroup$

As I have read the statement of the problem, neither of the given solutions is correct. Yours is flawed because you are assuming that the probability of observing an outcome is uniform over all possible outcomes. Either Brazil will win the World Cup, or they will not. You cannot assume that just because there are only two outcomes, that their chance to win must be equal to the chance they do not--otherwise, as you can see, such reasoning would lead to the conclusion that every nation would have a 50% chance of winning, which is absurd as there are many more than two nations playing.

Your friend's answer is also incorrect, but for a more subtle reason: his argument relies on the assumption that the two balls drawn are identifiable--that is to say, that there is a "first ball" and "second ball." Under such an assumption, the drawing of the balls can be thought to be sequential (but still without replacement), rather than simultaneous; and then the probability that the second ball is new given that the first ball is new, is equal to $5/9$ as he computed. But this is not what the problem stated: instead, it very clearly states that two balls were drawn, and one of them was observed to be new.

To get a better understanding of why this interpretation is not consistent with the way the question was posed, think of it this way: Suppose boys and girls are born with equal probability, and that the outcome of one birth is independent of any others. Furthermore, suppose I tell you that a given family has two children, at least one of which is a boy. What is the probability that the other child is also a boy? The answer is not $1/2$. There are four possible outcomes for any family with two children: $$\{(M,M), (M,F), (F,M), (F,F)\},$$ each of which has equal probability of $(1/2)^2 = 1/4$. But I said that this family has a son: so the only outcomes that you can choose from are $$\{(M,M), (M,F), (F,M)\}.$$ And these again occur with equal probability, but this time, the probability is $1/3$, not $1/4$. So the probability that the other child is male is $1/3$.

Now, back to your question: we can see that what we need to do is apply the boy/girl probability scenario to your new/old ball scenario: the only difference is in the probability distribution for old versus new balls. We have four outcomes for the two balls, with the following probability distribution: $$\begin{align*} \Pr[(N,N)] &= \frac{6}{10}\frac{5}{9} = \frac{1}{3}, \\ \Pr[(N,O)] &= \frac{6}{10}\frac{4}{9} = \frac{4}{15}, \\ \Pr[(O,N)] &= \frac{4}{10}\frac{6}{9} = \frac{4}{15}, \\ \Pr[(O,O)] &= \frac{4}{10}\frac{3}{9} = \frac{2}{15}. \end{align*} $$ But when we drew the balls, we saw at least one new ball. We did not see the $(O,O)$ case, which occurred with probability $2/15$. So the conditional probability is $$\Pr[(N,N) \mid \{(N,N) \vee (N,O) \vee (O,N)\}] = \frac{\Pr[(N,N)]}{1 - Pr[(O,O)]} = \frac{1/3}{1 - 2/15} = \frac{5}{13}.$$

$\endgroup$
0
$\begingroup$

The first answer is correct - it uses the definition of conditional probability to compute it.

The second answer is wrong - it (wrongly) assumes that the cases are equiprobable. E.g., imagine you are drawing from a bag with 99 red balls and 1 blue ball. According to the logic of the second answer, since there are two cases, the probability of drawing a blue ball is 1/2 and not 1%.

$\endgroup$
0
$\begingroup$

From your question and your two solutions I am not clear exactly what you are asking. So here are two answers, perhaps one of them solves your problem.

  1. Probability that the second ball is new, given that the first is new: if the first is new then the box contains $9$ balls, $5$ of them are new, and the probability that the second ball drawn is new is $5/9$. This matches your friend's answer, but as you can see there is actually no need to use conditional probability.
  2. Probability that both balls are new, given that at least one of them is new. We use the conditional probability formula: if $A$ is the event that both balls are new and $B$ is the event that at least one ball is new, then $$P(A|B)=\frac{P(A\cap B)}{P(B)}\ .$$ We have $$P(A\cap B)=P(A)=\frac{C(6,2)}{C(10,2)}=\frac{1}{3}$$ and $$P(B)=1-P(\hbox{both balls old})=1-\frac{C(4,2)}{C(10,2)}=\frac{13}{15}$$ so $$P(A|B)=\frac{5}{13}\ .$$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.