You flip a coin 4 times that has been weighted such that heads comes up twice as often as tails . What is the probability that all 4 of them are heads?

Probability of an event is the measure of its chance of occurrence. The probability of all 4 tossed coins in given context coming out as heads is 0.197 approximately.

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

[tex]X \sim B(n,p)[/tex]

The probability that out of n trials, there'd be x successes is given by

[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]

For the given case, let model the condition as:

Success = getting head on given biased coin

Probability of success = p = 2/3 (as head comes twice as often as tails, so probability of heads = twice probability of q = x say,

then 2x + x = 1(total probability is 1), or x = 1/3 = probability of tails,

thus, probability of heads= 2/3)

Failure = getting tail on given biased coin

Probability of failure = q = 1-p = 1-2/3 = 1/3

All coins' results are independent, thus, they are Bernoulli trials.

The count of Bernoulli trials is n = 4

Let random variable X tracks the number of heads obtained on tossing these 4 given biased coins.

Then,

[tex]X \sim B(4,2/3)[/tex]

The needed probability is

P(X = 4)

Using the probability function of binomial distribution, we get:
[tex]P(X = 4) = \: ^4C_4(2/3)^4(1/3)^0 = \dfrac{16}{81} \approx 0.197[/tex]

Thus, The probability of all 4 tossed coins in given context coming out as heads is 0.197 approximately.

Learn more about binomial distribution here:

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