the ratio of probability of 3 successes in 5 independent trials to the probability of 2 successes in 5 independent trials is 1/4.

What is the probability of 4 successes in 6 independent trials?

let prob of success be p

let prob of failure be q, where p+q = 1

Prob(3out of 5) = C(5,3) p^3 q^2 = 10p^3 q^2
prob(2 out of 5) = C(5,2)p^2 q^3 = 10 p^2 q^3

10p^3q^2/(10p^2q^3 = 1/4
q = 4p
in p+q=1
p + 4p=1
p = .2 , then p = .8

so prob(4 out of 6) = C(6,4) (.8)^4 (.2)^2
= .24576
or
= 768/3125

looks like I did not mind my p's and q's

from p+4p=1
p=.2, then q= .8

prob(4 out of 6 successes) = C(6,4) (.2)^4 (.8)^2
= .01536

p^3(1-p)^2*[5!/(2!*3!)] = P(3 out of 5)

p^2*(1-p)^3*[5!/(3!*2!)]= P(2 out of 5)

p/(1-p) = 1/4
p = 1/5 is the single-trial probability of success

P(4 out of 6) = p^5(1-p)^2*[6!/(4!*2!)]
= 3.2^10^-4*(0.64)*15 = 0.0307

To solve this problem, we can use the concept of binomial probabilities.

The probability mass function (PMF) for the binomial distribution is given by the formula:

P(X=k) = C(n, k) * p^k * (1-p)^(n-k)

Where:
P(X=k) is the probability of getting exactly k successes in n trials,
C(n, k) is the number of combinations of n items taken k at a time,
p is the probability of success in a single trial, and
(1-p) is the probability of failure in a single trial.

Let's break down the given information and solve the problem step by step:

1. The ratio of the probability of 3 successes in 5 independent trials to the probability of 2 successes in 5 independent trials is 1/4.

Let's denote the probability of success as p. We can set up the following equation:

P(X=3) / P(X=2) = 1/4

Using the PMF formula, we can write this equation as:

[C(5, 3) * p^3 * (1-p)^(5-3)] / [C(5, 2) * p^2 * (1-p)^(5-2)] = 1/4

2. Simplify the equation:

C(5, 3) = 5! / (3! * (5-3)!) = 10
C(5, 2) = 5! / (2! * (5-2)!) = 10

The equation becomes:

(10 * p^3 * (1-p)^2) / (10 * p^2 * (1-p)^3) = 1/4

Simplifying further:

(p^3 * (1-p)^2) / (p^2 * (1-p)^3) = 1/4

3. Cross-multiply and solve for p:

4 * (p^3 * (1-p)^2) = (p^2 * (1-p)^3)

4 * p^3 * (1-p)^2 = p^2 * (1-p)^3

Divide both sides by p^2:

4 * p * (1-p)^2 = (1-p)^3

Divide both sides by (1-p)^2:

4 * p = (1-p)

4p = 1 - p

5p = 1

p = 1/5

So, the probability of success in a single trial is 1/5.

4. Use the probability of success in a single trial to find the probability of 4 successes in 6 independent trials.

Using the PMF formula, we can plug in the values:

P(X=4) = C(6, 4) * (1/5)^4 * (1 - 1/5)^(6-4)

Simplifying:

C(6, 4) = 6! / (4! * (6-4)!) = 15

P(X=4) = 15 * (1/5)^4 * (4/5)^2

Calculating:

P(X=4) = 15 * (1/625) * (16/25)

P(X=4) = 15/15625

Therefore, the probability of 4 successes in 6 independent trials is 15/15625.