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I have the following problem from DeGroot:

A box contains 100 balls, of which 40 are red. Suppose that the balls are drawn from the box one at a time at random, without replacement. Determine (1) the probability that the first ball drawn will be red. (2) the probability that the fifteenth ball drawn will be red, and (3) the probability that the last ball drawn will be red.

For the first question:

  1. the probability that the first ball drawn will be red should be:

    $$\frac{40!}{(100-40)!}$$

  2. the probability that the fifteenth ball will be red:

    $$\frac{60! \times 40!}{(60-14)! \times (40-1)}$$

  3. the probability that the last ball drawn will be red:

    $$\frac{60! \times 40!}{100}$$

Am I on the right track with any of these?

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I am not sure that any of the answers is correct. For part 1, you only need to calculate the probability that a random chosen ball is red? That has a much simpler expression. (In fact, parts 2 and 3 also have the same answers if I am not mistaken.) –  Srivatsan Oct 14 '11 at 14:38
    
@Srivatsan Narayanan no, the question is about the probability that the FIRST ball is red –  Dbr Oct 14 '11 at 14:39
    
@SrivatsanNarayanan You are not mistaken: all three probabilities have the same value. This is a typical source of confusion for beginning students: whether we use sampling without replacement or sampling with replacement, the probability of drawing a red ball on the $n$-th draw is the same for all $n$. The statement is obvious for sampling with replacement but not so obvious for sampling without replacement as in this case. –  Dilip Sarwate Oct 14 '11 at 15:02
    
To add to @Dilip's point, there's no difference between sampling with and without replacement as long as you consider the $n$-th draw for some single $n$. But the two distributions start to behave differently as soon as you consider the $n$-th and $m$-th draws. For example, the probability that the $10$-th ball is red and $12$-th ball is not red is different in the two cases. –  Srivatsan Oct 14 '11 at 15:11
1  
Suppose Prof. deGroot told you to draw just one ball instead of all $100$ one after another. What would be the probability that it would be red? And would this probability (of the first ball being red) change if, after you drew the first ball, he said "I changed my mind; now draw all the remaining balls one after another without replacement."? –  Dilip Sarwate Oct 14 '11 at 15:11

2 Answers 2

up vote 6 down vote accepted

The answers are (a) $40/100$; (b) $40/100$; (c) $40/100$.

(a) Since balls tend to roll around, let us imagine instead that we have $100$ cards, with the numbers $1$ to $100$ written on them. The cards with numbers $1$ to $40$ are red, and the rest are blue.

The cards are shuffled thoroughly, and we deal the top card. There are $100$ cards, and each of them is equally likely to be the top card. Since $40$ of the cards are red, it follows that the probability of "success" (first card drawn is red) is $40/100$.

(b) If we think about (b) the right way, it should be equally clear that the probability is $40/100$. The probability that any particular card is the fifteenth one drawn is the same as the probability that it is the first one drawn: all permutations of the $100$ cards are equally likely. It follows that the probability that the fifteenth card drawn is red is $40/100$.

(c) First, fifteenth, eighty-eighth, last, it is all the same, since all permutations of the cards are equally likely.

Another way: We look again at (a), using a more complicated sample space. Again we use the card analogy. There are $\binom{100}{40}$ ways to decide on the locations of the $40$ red cards (but not which red cards occupy these locations). All these ways are equally likely.

In how many ways can we choose these $40$ locations so that Position $1$ is one of the chosen locations? We only need to choose $39$ locations, and this can be done in $\binom{99}{39}$ ways. So the desired probability is $$\frac{\binom{99}{39}}{\binom{100}{40}}.$$ Compute. The top is $\frac{99!}{39!60!}$ and the bottom is $\frac{100!}{40!60!}$. Divide. There is a lot of cancellation. We quickly get $40/100$.

Now count the number of ways to find locations for the reds that include Position $15$. An identical argument shows that the number of such choices is $\binom{99}{39}$, so again we get probability $40/100$.

Another other way: There are other reasonable choices of sample space. The largest natural one is the set of $100!$ permutations of our $100$ cards. Now we count the permutations that have a red in a certain position, say Position $15$.

The red in Position $15$ can be chosen in $40$ ways. For each of these ways, the remaining $99$ cards can be arranged in $99!$ ways, for a total of $(40)(99!)$. Thus the probability that we have a red in Position $15$ is $$\frac{(40)(99!)}{100!},$$ which quickly simplifies to $40/100$. The same argument works for any other specific position.

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I find it instructive to lead students through the horrible, brute force calculation before teaching them that the end result is best understood using symmetry.

Let's calculate the probability that the 15th ball is red, taking into account all the balls drawn previously. For example, one of the outcomes that make up our event is $$RNNRNNNRRRNRNRR$$ where $R$ means a red ball and $N$ a non-red ball. The probability of getting this particular outcome is $${40\over 100}\cdot{60\over 99}\cdot{59\over 98}\cdots {33\over 86}={(40)_8 \, (60)_7\over (100)_{15}}.$$ We simplify the product of fractions using the Pochhammer symbol. The "7" and "8" are the number of $N$s and $R$s in the outcome.

Adding the probabilities of all such outcomes gives $$\mathbb{P}(\mbox{15th ball is red})={1\over (100)_{15}}\sum_k {14\choose k} (40)_{15-k}\ (60)_k $$ $$={1\over (100)_{15}}\ 40\ \sum_k{14\choose k} (39)_{14-k}\ (60)_k ={1\over (100)_{15}}\ 40\ (99)_{14} ={40\over 100}.$$ Amazing!

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