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I am working on some homework questions and I am done everything except this one last question which I am stuck at. I am uncertain of how I can explain or find the probability in this case. I had a similar question where I just tossed a normal coin and they asked the probability to get heads or tails and that was relatively easy but with this one I am not sure..I have tried reading some other coin toss probability questions on this forum but they arent similar to this at all.

Firstly I dont even understand the question because of how complicated it is. All I understand is there are 1000 "items" with 11 "items" inside each of these 1000 items. Then I am clueless.

After your explanation Ross I was able to do 5i) and I think I did it right I am unsure on induction as that is my weak point.

5i) This question in a simpler form is asking what is the chance that all the coin tosses turn out heads or all the coin tosses turn out tails for a set A1. This means (1/2) (1/2)(1/2) (1/2)(1/2) (1/2) (1/2)(1/2) (1/2)(1/2) (1/2) which is (1/2)^11. This means the probability to get heads 11 times in a row is (1/2)^11. Now there is also a probability of our interest that all the coins turn out tails. The probability that all coin tosses turn out tails is exactly the same as the probability that all coin tosses turn out heads. Which means our probability to have ALL heads or ALL tails is all heads OR all tails. Namely, (1/2)^11+(1/2)^11 is our probability to have all red or all blue since heads and tails correspond to the colors.

  • $\begingroup$ For i) the first toss can be either heads or tails, so the probability is $(\frac 12)^{10}$ Can you prove ii for $n=2$? $\endgroup$ Commented Mar 19, 2012 at 14:14
  • $\begingroup$ Firstly I dont even understand the question because of how complicated it is... You mean, even part (i) is too complicated? $\endgroup$
    – Did
    Commented Mar 19, 2012 at 16:50

2 Answers 2


Your answer to (i) is fine, though you should finish it off: the probability that $A_1$ is monochromatic is $$\left(\frac12\right)^{11}+\left(\frac12\right)^{11}=\frac2{2048}=\frac1{1024}\;.$$

If you’re having trouble with induction, I suggest that you skip (ii) for the moment, simply assuming that the stated result is true, and go on to try (iii), (iv), and (v).

For (iii), $M_i$ is the event that $A_i$ is monochromatic, so $M_1\cup M_2\cup\dots\cup M_{1000}$ is what event? You used the right idea at the end of your solution to (i).

For (iv), remember that we’re assuming for the moment that (ii) is true. In particular, we’re assuming that $$\operatorname{Pr}(M_1\cup M_2\cup\dots\cup M_{1000})\le\sum_{i=1}^{1000}\operatorname{Pr}(M_i)\;.\tag{1}$$ Use your answer to (i) to evaluate the righthand side of $(1)$. Now, what event has probability $$1-\operatorname{Pr}(M_1\cup M_2\cup\dots\cup M_{1000})\;?$$

For (v), you just need to realize that if something has a positive probability of occurring at random, it cannot be impossible.

Now back to (ii). You want to prove by induction on $n$ that $$\operatorname{Pr}(M_1\cup M_2\cup\dots\cup M_n)\le\sum_{i=1}^n\operatorname{Pr}(M_i)\;.\tag{2}$$ Is $(2)$ true when $n=1$? Sure: in that case it just says that $\operatorname{Pr}(M_1)=\operatorname{Pr}(M_1)$. What about for $n=2$? In that case $(2)$ says that $$\operatorname{Pr}(M_1\cup M_2)\le\operatorname{Pr}(M_1)+\operatorname{Pr}(M_2)\;;\tag{3}$$ if you can explain why that’s true, you have everything that you need for the rest of the argument.

Now for the induction step you want to suppose that $(2)$ is true and somehow use that to conclude that the corresponding statement for $n+1$ is true, i.e, that $$\operatorname{Pr}(M_1\cup M_2\cup\dots\cup M_{n+1})\le\sum_{i=1}^{n+1}\operatorname{Pr}(M_i)\;.\tag{4}$$ Think of $M_1\cup M_2\cup\dots\cup M_{n+1}$ as $(M_1\cup M_2\cup\dots\cup M_n)\cup M_{n+1}$. In fact, to reduce the notational clutter, let’s let $M=M_1\cup M_2\cup\dots\cup M_n$ and rewrite $(4)$ as $$\operatorname{Pr}(M\cup M_{n+1})\le\sum_{i=1}^n\operatorname{Pr}(M_i)+\operatorname{Pr}(M_{n+1})\;.\tag{5}$$ Remember that we’re assuming $(2)$ as our induction hypothesis, and $(2)$ can be rewritten as $$\operatorname{Pr}(M)\le\sum_{i=1}^n\operatorname{Pr}(M_i)\;;$$ can you see how this makes proving $(5)$ pretty nearly the same as proving $(2)$?


For i), imagine the set is $\{1,2,3,4,5,6,7,8,9,10,11\}$. You flip a coin to color each number. What is the chance that all the numbers get the same color?

For ii) for $n=2,\ $ Pr($M_1 \cup M_2)$ is the chance that either event $M_1$ or event $M_2$ or both happen. Think of a Venn diagram-it is the area inside either circle.

See if this helps you along.


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