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Find the probability of a success for the binomial distribution satisfying the relation 4P(x=4)= P(x=2) and having the parameter n as 6?
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Find the probability of a success for the binomial distribution satisf...
Binomial Distribution with Parameter n=6

The binomial distribution is a probability distribution that describes the number of successes in a fixed number of independent trials, where each trial has the same probability of success.

In this case, the parameter n=6 means that there are 6 independent trials.

Relation between P(x=4) and P(x=2)

The given relation 4P(x=4)= P(x=2) means that the probability of getting 4 successes in 6 trials is four times the probability of getting 2 successes in 6 trials.

Using the Binomial Distribution Formula

We can use the binomial distribution formula to find the probability of a success, denoted by p.

The binomial distribution formula is P(x=k) = (n choose k) p^k (1-p)^(n-k), where (n choose k) is the binomial coefficient.

Substituting the given values, we get:

4*(6 choose 4) p^4 (1-p)^2 = (6 choose 2) p^2 (1-p)^4

Simplifying the equation, we get:

15p^2(1-p)^2 = p^4(1-p)^4

15(1-p) = p^2

15 - 15p = p^2

p^2 + 15p - 15 = 0

Solving for p using the quadratic formula, we get:

p = (-15 ± sqrt(225 + 60)) / 2

p = (-15 ± sqrt(285)) / 2

Since the probability of a success cannot be negative, we take the positive root:

p = (-15 + sqrt(285)) / 2

p ≈ 0.258

Therefore, the probability of a success is approximately 0.258.
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Find the probability of a success for the binomial distribution satisfying the relation 4P(x=4)= P(x=2) and having the parameter n as 6?
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