How to tell whether a probability is conditional?

Question

In a class of $32$ children, $18$ are Chinese, $8$ are Malay, $4$ are Indian and $2$ are Eurasian. One child is chosen at random from the class. Another child is then chosen at random from the remaining children. A third child is then chosen from the remaining children. Find the probability that the three children chosen are of different races, given that the second child is Chinese.

I am confused about the wording of the question. My confusion is on whether I should use conditional probability for this problem.

When the problem statement says "Find the probability of A given B", I always think of conditional probability, in which case $$P(\text{different races given second Chinese})= \dfrac{P(\text{different races and second Chinese})}{P(\text{second Chinese})}\\= \dfrac{(21/310)}{(9/16)}=\dfrac{56}{465}.$$

However couldn't the question be equivalently rephrased as follows? "Find the probability of picking three children of different races where the second child is Chinese." In this case the probability is just $$P(\text{different races and second Chinese})= \dfrac{21}{310}.$$

This question was part of a recent probability test in which I lost marks for giving the first answer (56/465). I would like clarification on whether my conception is correct.

Answer 1

I don't think that it is a coincidence that my answer agrees with the OP's first computation. So, although the OP didn't actually describe his thinking, I went ahead and posted my answer.

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Yes, you do want to use conditional probability, but (in this instance) the application of it is convoluted.

Let $~A~$ denote the event that the $~3~$ children are of three different races.

Let $~B~$ denote the event that the 2nd child is Chinese.

Then, you are asked for

$$p(A ~| ~B ~) = \frac{p(A \cap B)}{p(B)}. \tag1 $$

So, the entire problem is reduced to computing the numerator and denominator, in (1) above. Since this problem is a combinatorics-oriented problem, I will express the probabilities of $~p(A \cap B)~$ and $~p(B)~$ combinatorically.

The denominator is easy. There are $~\displaystyle \binom{32}{1}~$ ways of selecting the 2nd student, of which $~\displaystyle \binom{18}{1}~$ of these ways will result in the person being Chinese.

Therefore,

$$p(B) = \frac{\binom{18}{1}}{\binom{32}{1}} = \frac{18}{32}.$$

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To express the $~p( ~A \cap ~B ~) ~$ as $~\dfrac{N}{D},~$ the simplest approach is to assume that order of selection is relevant in both the computation of $~N~$ and the computation of $~D.$

So, $~D = \dfrac{32!}{(32 - 3)!} = 32 \times 31 \times 30.$

Computation of $~N,~$ the numerator is convoluted, and must be broken down into the following mutually exclusive cases, that must then be summed together:

• Malay, Chinese, Indian.
  $~\displaystyle \binom{8}{1} \times \binom{18}{1} \times \binom{4}{1} = 8 \times 18 \times 4.$

• Malay, Chinese, Eurasian.
  Similar to the previous bullet point, the computation is $~8 \times 18 \times 2.$

• Indian, Chinese, Malay.
  $4 \times 18 \times 8.$

• Indian, Chinese, Eurasian.
  $4 \times 18 \times 2.$

• Eurasian, Chinese, Malay.
  $2 \times 18 \times 8.$

• Eurasian, Chinese, Indian.
  $2 \times 18 \times 4.$

Therefore,

$$N = [ ~8 \times 18 \times 4 ~] + [ ~8 \times 18 \times 2 ~] + [ ~4 \times 18 \times 8 ~] + [ ~4 \times 18 \times 2 ~] \\ + [ ~2 \times 18 \times 8 ~] + [ ~2 \times 18 \times 4 ~]. $$

Then, $~N~$ can be re-expressed as

$$N = 18 \times \left\{ ~[ ~8 \times 4 ~] + [ ~8 \times 2 ~] + [ ~4 \times 8 ~] + [ ~4 \times 2 ~] + [ ~2 \times 8 ~] + [ ~2 \times 4 ~] ~\right\} $$

$$= 18 \times 112.$$

Therefore

$$p(A \cap B) = \frac{N}{D} = \frac{18 \times 112}{32 \times 31 \times 30}.$$

Therefore,

$$\frac{p(A \cap B)}{p(B)} = \frac{\frac{18 \times 112}{32 \times 31 \times 30}}{\frac{18}{32}}. \tag1 $$

Notice that in (1) above, both the numerator and the denominator have a common fraction of $~\dfrac{18}{32}.$

Cancelling this out gives

$$\frac{p(A \cap B)}{p(B)} = \frac{\frac{112}{31 \times 30}}{1} = \frac{112}{31 \times 30} = \frac{56}{465}.$$

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$~\underline{\text{Addendum}}$

Hijacking the analysis in one of the comments of lulu, the problem can be re-interpreted as a purely probability problem that does not involve conditional probability.

Since the 2nd child chosen is Chinese, as far as the selection of the 1st and 3rd children are concerned, you are now working in a universe of $~31~$ people, of which the number of Malay, Indian, and Eurasian are $~8, ~4,~$ and $~2~$ respectively.

So, the probability can be expressed as

$$\dfrac{N}{D} ~: ~D = 31 \times 30,$$

where

$$N = [ ~8 \times (4 + 2) ~] + [ ~4 \times (8 + 2) ~] + [ ~2 \times (8 + 4) ~] = 112.$$

So,

$$\frac{N}{D} = \frac{112}{31 \times 30} = \frac{56}{465}.$$

Answer 2

Find the probability that the three children chosen are of different races, given that the second child is Chinese.

$$P(\text{different races }\textbf{given}\text{ second Chinese})=\dfrac{56}{465}.$$

This is correct. Equivalently: If the second chosen child is Chinese, find the probability that the three children chosen are of different races.

However couldn't the question be equivalently rephrased as follows? "Find the probability of picking three children of different races where the second child is Chinese." In this case the probability is just $$P(\text{different races }\textbf{and}\text{ second Chinese})= \dfrac{21}{310}.$$

This new question means Find the probability that the three children chosen are of different races such that the second child is Chinese (the word "where" has been replaced without altering meaning), and its correct answer is indeed $\frac{21}{310}.$ So, it is not equivalent to the original question.

Answer 3

Order doesn't matter. We can consider that this chinese child appears in 1st position.

After this chinese child, what is the probability that child n°2 is not chinese ? $\frac{14}{31}$

But if we start like this, we will not be able to compute probability that child n°3 is ok.

Probability that child n°2 is Malay is $\frac{8}{31}$, and in this case, proba of success is $\frac{6}{30}$

Probability that child n°2 is Indian is $\frac{4}{31}$, and in this case, proba of success is $\frac{10}{30}$

Probability that child n°2 is Eurasian is $\frac{2}{31}$, and in this case, proba of success is $\frac{12}{30}$

Total proba of success is $\dfrac{8 \times 6 + 4 \times 10 + 2 \times 12}{31 \times 30}= \dfrac{112}{31\times 30}$

As far as I remember, all problems with conditional probability can also be solved with simple probability. We just need to ajust the universe.