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according to my previous link probability calculation in dice throwing i have tried to solve following problem:

A bag contains $4$ white and $3$ black balls. Four balls are successively drawn out with replacement. Find the probability that they are alternatively of different colors.

so first thing ,we have in total $7$ ball,then with replacement means that we always have $7$ ball,so we are taking and putting again in,if we want that four ball occurs in such order,so that create alternative colors would be

black,white,black,white

or

white,black,white,black

this order is asked right?now my question is what is a number of other possible occurring? for example one possible thing is all of them white,or all of them black,also for each white we have $4$ possibility or $12$ and again for each black we have $3$ possibility again $12$,so in total it would be $26$ right?i added all black and all white?or just $24$,first of all what i have tried was

$4/7$ * $3/7$ * $4/7$ * $3/7$ which is equal to $ 144/2401$,but this answer is wrong and that why i would like to count total number of let say success and divide it by total number of occurring of balls.please help me

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dato datuashvili
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2 Answers2

2

Your calculation of

$$\frac47\cdot\frac37\cdot\frac47\cdot\frac37=\frac{144}{2401}$$

gives the probability of drawing the sequence BWBW, where B stands for black and W for white. That much is okay, but you also need to calculate the probability of the sequence WBWB, since it also satisfies the condition that the colors alternate. Its probability is the same:

$$\frac37\cdot\frac47\cdot\frac37\cdot\frac47=\frac{144}{2401}\;.$$

Thus, the probability of getting a sequence in which the colors alternate is

$$\frac{144}{2401}+\frac{144}{2401}=\frac{288}{2401}\;.$$

In this problem there are $2^4=16$ possible sequences of colors, but they don’t all have the same probability. For instance, the probability of WWWW is $\left(\frac47\right)^4=\frac{256}{2401}$, while the probability of BBBB is only $\left(\frac37\right)^4=\frac{81}{2401}$. Thus, counting outcomes is not a good way to approach this problem.

Brian M. Scott
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  • aaaa ui,i see,thanks very much,i have just mentioned that there is two alternative,but i had not counted probability for second one – dato datuashvili Aug 01 '13 at 08:34
  • @dato: You’re welcome! – Brian M. Scott Aug 01 '13 at 08:35
  • thanks very much @Brian M. Scott you are helping me too much,if i have passed GRE and then TOEFEL ,i will study without payment in PHD degree,i have to take $700$ from $800$ or answer $25$ from $40$ – dato datuashvili Aug 01 '13 at 08:39
  • Wouldn't the probability of BWBW be (4/7) * (3/6) * (3/5) * (2/4) as each case is not independent. If I take one ball out, the sample space is down by 1 too. Same for WBWB – Hemil Jun 10 '22 at 13:33
  • @Hemil: No, because we are told that the balls are drawn with replacement: each ball that is drawn is returned to the bag before the next ball is drawn. – Brian M. Scott Jun 10 '22 at 20:01
  • @BrianM.Scott ah sorry. I actually had a similar question for without replacement. I misread the question. Thanks a lot for replying on a decade old question though – Hemil Jun 11 '22 at 13:38
  • @Hemil: You’re welcome. – Brian M. Scott Jun 11 '22 at 17:42
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First of all, the reason why $\frac47 \times \frac37 \times \frac47 \times \frac37$ was wrong is simple - this only accounts for the black-white-black-white case. As you said, there are two ways for colors to be alternating - black-white-black-white, and white-black-white-black. The probability of white-black-white-black is $\frac37 \times \frac47 \times \frac37 \times \frac 47$. Adding these two, the total probability of alternating colors is $\frac{288}{2401}$.

It also seems like there's a larger issue with understanding here. Perhaps it would be easier to think of each ball as having its own unique identification - let the black balls be $B_1, B_2, B_3$ and $B_4$. Similarly, let the whites be $W_1, W_2$, and $W_3$.

Now, you proposed counting the "successes". A success would be come in the form of black-white-black-white or white-black-white-black. The total number of black-white-black-white is $4 \times 3 \times 4 \times 3$, as the blacks can be any of the $B_i$, and the whites can be any of the $W_j$, where $i = 1, 2, 3, 4$ and $j = 1, 2, 3$. Similarly, the total number of white-black-white-black would be $3 \times 4 \times 3 \times 4$. This makes a total of $144 + 144 = 288$ successes.

Counting the total number of ball combinations is actually quite a bit easier! Each ball can be any of $B_1, B_2, B_3, B_4, W_1, W_2$, or $W_3$. This means that there are $7$ options for each ball, for a total of $7 \times 7 \times 7 \times 7 = 2401$ ball combinations.

Dividing the successes by the total, we have $\frac{288}{2401}$. You'll notice that these two methods are the exact same thing - it's a question of how you'd like to think about it.

Cheers!

HJ32
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