0

I'm having a hard time visualizing this question:

Assume you flip $3$ coins. What is the probability that you get exactly one head.

$P(\text{head}) = \frac{1}{2}$. $P(\text{Total outcomes}) = 2^3$

I was told in combinations order doesn't matter. The answer is $\frac{{3\choose1}}{2^3}$. Which would be

$\{HTT, TTH, THT\}$. Order matters doesn't it?

hello
  • 17
  • 1
    You are correct. In combinations, order does not matter. In permutations, it does. – Tony Hellmuth Jul 01 '18 at 06:01
  • Order does matter in this problem. The number of favorable cases is $\binom{3}{1}$ since we are choosing which of the three potential positions for a head is filled with heads. Similarly, the probability of exactly two heads in four tosses of a fair coin is $\frac{\binom{4}{2}}{2^4}$ since we must choose which two of the four positions in the sequence of heads and tails will be filled with heads. – N. F. Taussig Jul 01 '18 at 08:09

2 Answers2

1

Yes, order matters: imagine the analogous problem with just two coin flips.

The three possible outcomes are: two heads, two tails, one of each; if we mistakenly discount the importance of order, we might mistakenly believe each of these three scenarios is equally likely to occur. In other words, that each of the three scenarios occurs with probability $1/3$.

But, try this experiment - flipping a coin twice - many times. See what you observe to be the probability of these scenarios. Over time, it should look like two heads, $1/4$; two tails, $1/4$; one of each, $1/2$. This is because by correctly taking order into account, there are the four scenarios of HH; TT; HT, TH.

A similar line of reasoning applies to the problem about which you ask.

Benjamin Dickman
  • 14,451
0

You can see it this way.

HHH- 1

HHT- $\frac{3!}{2!}$ = 3

HTT- $\frac{3!}{2!}$ = 3

TTT- 1

Since the question asks for exaxtly one head, the probability would be $\frac{3}{8}$

You are using combinatorics to find out the possible combinations of HHH, HHT, HTT and TTT. That is why orders matter in this case. Sorry, I don't know how to word it properly.

  • I think OP may want distinction between order and unordered in considering similar questions. – Tony Hellmuth Jul 01 '18 at 06:42
  • @TonyHellmuth I don't understand how to make it more clear. What I want to say is he is using combinatorics to find out the possible combinations of HHH, HHT, HTT and TTT. That is why orders matter in this case. Can you help me word it properly then? –  Jul 01 '18 at 06:48
  • It is your answer and a fine one too :) Just maybe add a few points on the actual question "Order matters doesn't it?" – Tony Hellmuth Jul 01 '18 at 06:50
  • @TonyHellmuth thank you. I have edited it. –  Jul 01 '18 at 06:52