I have actually done probability with coins before, however this concern stumped me. How? due to the fact that we only have ONE coin, and we don"t recognize how plenty of times the coin is tossed. I recognize that with one coin, the probability of getting a head is 1. And the variety of outcomes is 2. However, ns don"t recognize the following step after this, specifically when I"m not offered information on how countless times the coin must be tossed.
You are watching: What is the probability of getting tails 4 times in a row when you flip a coin?
Any aid would be great. Or maybe simply a tip on looking in ~ this difficulty from a various perspective? say thanks to you!
I"ve also started statistics together well.
How ns look into this concern is the other way around:
Rather than in search of 4 in a row, i look at the probability of not having 4 top in a heat (having the compliment of the probability).
Let speak P(A) = having at least 4 heads before very 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") $
$1-P(T)+P(H\cap T)+P(H\cap H \cap T)+P(H \cap H \cap H \cap T)$
You desire the probability that flipping in ~ least four heads before obtaining the very first tail.
Thus you desire the probability the at least the an initial four tosses space heads.
This is $1/2^4$.
The probability of getting a heads an initial is 1/2.The probability of getting 2 heads in a heat is 1/2 the that, or 1/4.The probability of gaining 3 heads in a heat is 1/2 that that, or 1/8.The probability of obtaining 4 top in a heat is 1/2 of that, or 1/16.
After that... It doesn"t matter... You have actually at the very least 4 heads.
The price is 1/16.
There room several feasible approaches.
Getting 4 top to start has actually probability $(1/2)^4 = 1/16$ together inthe comment by
More formally, outcomes that satisfy your problem areHHHHT, HHHHHT, HHHHHHT, etc. So the total probabilityis 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, friend cannote that $(1/2)A = (1/2)^6 + 1/2)^7 + \dots,$ therefore that$A - (1/2)A = = (1/2)A = (1/2)^5$ and $A = (1/2)^4,$ i m sorry is the very same as the previous answer.
I perform not understand why
manmood (+1) that appeared while was inputting this.I hope one of these explanations is clear to you. The an essential points throughoutis that we"re presume the coin is fair <$P(H) = 1/2$> and that the tosses room independent.
Also, if your book includes the geometric distribution, you must lookat that due to the fact that it is related to this problem. I don"t desire to discussthe geometric circulation in this Answer due to the fact that there room at leasttwo execution of it, and discussing the wrong one might be confusing.
Well, h=heads, t=tailsSample space=(hhhh),(hhht),(hhth),(hhtt),(hthh),(htht),(htth),(httt), (thhh),(thht),(thth),(thtt),(tthh),(ttht),(ttth),(tttt)There are 16 total outcomes, and also only 1 of these outcomes outcomes in 4 heads. This means the probability of landing every 4 heads in 4 tosses is 1 the end of the 16.So the prize is 1/16.As for the "before flipping a tail", that doesn"t seem to matter because no issue the outcome there is still only 1 in 16 chances to get 4 top flipped.
The Probability of tossing 4 heads is a row is usually the multiplication of the probability of each toss that a head i m sorry is 1/2
P(H,H,H,H) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16
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