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

I have done probability with coins before, yet this inquiry stumped me. How? since we only have actually **ONE coin**, and also we don"t know how numerous times the coin is tossed. I know that through one coin, the probability of getting a head is 1. And the number of outcomes is 2. However, ns don"t understand the following step ~ this, especially when I"m not offered information top top how countless times the coin need to be tossed.

You are watching: Probability of getting heads 4 times in a row

Any assist would be great. Or maybe simply a reminder on looking at this trouble from a different perspective? give thanks to you!

I"ve likewise started statistics together well.

How i look right into this inquiry is the other method around:

Rather than looking for 4 in a row, ns look in ~ the probability of not having actually 4 top in a row (having the compliment that the probability).

Let say P(A) = having actually at least 4 heads before first 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 desire the probability of flipping at least four heads prior to obtaining the first tail.

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

This is $1/2^4$.

The probability of obtaining a heads very first is 1/2.The probability of acquiring 2 heads in a row is 1/2 of that, or 1/4.The probability of gaining 3 top in a row is 1/2 that that, or 1/8.The probability of getting 4 top in a heat is 1/2 of that, or 1/16.

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

The price is 1/16.

There space several possible approaches.

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

DonatPants.

More formally, outcomes that satisfy your condition areHHHHT, HHHHHT, HHHHHHT, etc. Therefore the complete probabilityis the geometric collection 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 recognize it, friend cannote the $(1/2)A = (1/2)^6 + 1/2)^7 + dots,$ so that$A - (1/2)A = = (1/2)A = (1/2)^5$ and also $A = (1/2)^4,$ i m sorry is the very same as the vault answer.

I execute not understand why

manmood (+1) that appeared while was typing this.I hope one of these explanations is clear to you. The crucial points throughoutis the we"re presume the coin is fair <$P(H) = 1/2$> and that the tosses are independent.

Also, if your publication includes the geometric distribution, you should lookat that because it is regarded this problem. Ns don"t want to discussthe geometric distribution in this Answer due to the fact that there room at leasttwo version of it, and also 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 room 16 full outcomes, and also only 1 of these outcomes results in 4 heads. This means the probability the landing every 4 top in 4 tosses is 1 out of the 16.So the prize is 1/16.As because that the "before flipping a tail", it doesn"t seem come matter since no matter the outcome there is still just 1 in 16 possibilities to obtain 4 heads flipped.

The Probability of tossing 4 heads is a heat is basically the multiplication of the probability of every toss of a head i beg your pardon is 1/2

P(H,H,H,H) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16

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Probability of picking a biased coin $C$ which has actually probability $3/15$ of gaining heads, assuming we obtained head top top the first toss

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