I am stuck on a problem for my stats class. I have not done any work but I will explain why. Here is the problem;
- A coin is flipped until you get a tails. What is the probability of getting at least 4 heads?
I have done probability with coins before, but this question stumped me. How? Because we only have ONE coin, and we don't know how many times the coin is tossed. I know that with one coin, the probability of getting a head is 1. And the number of outcomes is 2. However, I don't know the next step after this, especially when I'm not given information on how many times the coin should be tossed.
Any help would be great. Or maybe just a tip on looking at this problem from a different perspective? Thank you!
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$\begingroup$I've also started statistics as well.
How I look into this question is the other way around:
Rather than looking for 4 in a row, I look at the probability of not having 4 heads in a row (having the compliment of the probability).
Let say P(A) = Having at least 4 heads before first tail
$P(A') = P(T)+P(H\cap T)+P(H\cap H \cap T)+P(H \cap H \cap H \cap T)$
$P(A')+P(A) = 1 \rightarrow P(A) = 1-P(A') $
$1-P(T)+P(H\cap T)+P(H\cap H \cap T)+P(H \cap H \cap H \cap T)$
$ 1-[1/2 + 1/2^2 + 1/2^3 +1/2^4] = 1-15/16 = 1/16 $
$\endgroup$ $\begingroup$You want the probability of flipping at least four heads before obtaining the first tail.
Thus you want the probability that at least the first four tosses are heads.
This is $1/2^4$.
$\endgroup$ $\begingroup$The probability of getting a heads first is 1/2. The probability of getting 2 heads in a row is 1/2 of that, or 1/4. The probability of getting 3 heads in a row is 1/2 of that, or 1/8. The probability of getting 4 heads in a row is 1/2 of that, or 1/16.
After that... it doesn't matter... you have at least 4 heads.
The answer is 1/16.
$\endgroup$ 2 $\begingroup$There are several possible approaches.
Getting 4 heads to start has probability $(1/2)^4 = 1/16$ as in the comment by @DonatPants.
More formally, outcomes that satisfy your condition areHHHHT, HHHHHT, HHHHHHT, etc. So the total probability
is the geometric series with probability
$A = (1/2)^5 + (1/2)^6 + (1/2)^7 + \dots .$
There is a formula for summing this series. If you don't know it, you can note that $(1/2)A = (1/2)^6 + 1/2)^7 + \dots,$ so that $A - (1/2)A = = (1/2)A = (1/2)^5$ and $A = (1/2)^4,$ which is the same as the previous answer.
I do not know why @MarkusStuhr has withdrawn his Answer. The explanation by @Joel (+1) as does the answer by @manmood (+1) that appeared while was typing this. I hope one of these explanations is clear to you. The key points throughout is that we're assuming the coin is fair [$P(H) = 1/2$] and that the tosses are independent.
Also, if your book includes the geometric distribution, you should look at that because it is related to this problem. I don't want to discuss the geometric distribution in this Answer because there are at least two versions of it, and discussing the wrong one might be confusing.
$\endgroup$ $\begingroup$Well, h=heads, t=tails Sample space={(hhhh),(hhht),(hhth),(hhtt),(hthh),(htht),(htth),(httt), (thhh),(thht),(thth),(thtt),(tthh),(ttht),(ttth),(tttt)} There are 16 total outcomes, and only 1 of these outcomes results in 4 heads. This means the probability of landing all 4 heads in 4 tosses is 1 out of the 16. So the answer is 1/16. As for the "before flipping a tail", it doesn't seem to matter because no matter the outcome there is still only 1 in 16 chances to get 4 heads flipped.
$\endgroup$ $\begingroup$The Probability of tossing 4 heads is a row is basically the multiplication of the probability of each toss of a head which is 1/2
P(H,H,H,H) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
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