*A coin is flipped until you acquire a tails. What is the probability ofgetting at least 4 heads?*

I have actually done probability with coins before, yet this concern stumped me. How? due to the fact that we only have **ONE coin**, and also we don"t know how plenty of times the coin is tossed. I recognize that through one coin, the probability of obtaining a head is 1. And also the number of outcomes is 2. However, ns don"t know the following step after this, especially when I"m not given information top top how many times the coin need to be tossed.

You are watching: Probability of getting heads 3 times in a row

Any help would be great. Or maybe simply a reminder on looking in ~ this trouble from a various perspective? give thanks to you!

I"ve also started statistics as well.

How i look into this question is the other means around:

Rather than searching for 4 in a row, i look at the probability the not having 4 top in a row (having the compliment that the probability).

Let say P(A) = having at least 4 top before an initial tail

$P(A") = P(T)+P(Hcap T)+P(Hcap H cap T)+P(H cap H cap H cap T)$

$P(A")+P(A) = 1 ightarrow P(A) = 1-P(A") $

$1-P(T)+P(Hcap T)+P(Hcap H cap T)+P(H cap H cap H cap T)$

You want the probability the flipping in ~ least 4 heads before obtaining the an initial tail.

Thus you want the probability that at the very least the an initial four tosses are heads.

This is $1/2^4$.

The probability of getting a heads an initial is 1/2.The probability of acquiring 2 top in a heat is 1/2 of that, or 1/4.The probability of obtaining 3 top in a heat is 1/2 the that, or 1/8.The probability of obtaining 4 heads in a heat is 1/2 of that, or 1/16.

After that... It doesn"t matter... You have actually at least 4 heads.

The prize is 1/16.

There space several feasible approaches.

Getting 4 heads to start has probability $(1/2)^4 = 1/16$ together inthe comment through

DonatPants.

More formally, outcomes that fulfill your problem areHHHHT, HHHHHT, HHHHHHT, etc. So the full 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, you cannote that $(1/2)A = (1/2)^6 + 1/2)^7 + dots,$ for this reason that$A - (1/2)A = = (1/2)A = (1/2)^5$ and also $A = (1/2)^4,$ i beg your pardon is the same as the previous answer.

I execute not know why

manmood (+1) that appeared while was inputting this.I hope among these explanations is clean to you. The an essential points throughoutis the we"re assuming the coin is fair <$P(H) = 1/2$> and that the tosses room independent.

Also, if your book includes the geometric distribution, you have to lookat that since it is regarded this problem. Ns don"t want to discussthe geometric circulation in this Answer because there space at leasttwo version of it, and discussing the dorn one can 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 full outcomes, and only 1 of this 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 answer is 1/16.As for the "before flipping a tail", the doesn"t seem to matter because no matter the outcome over there is still only 1 in 16 possibilities to gain 4 heads flipped.

The Probability of tossing 4 heads is a row is basically the multiplication of the probability of each toss the 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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