An experiment succeeds twice as often as it fails. Find the probability that in the next six trails, there will be at least 4 successes.

To solve this problem, we will use the binomial probability formula. The binomial probability formula is used to find the probability of exactly k successes in n trials, where the probability of success on any given trial is p, and the probability of failure is q = 1 - p.

Step 1: Understanding the Problem
We are given that the experiment succeeds twice as often as it fails. This means the probability of success (p) is twice the probability of failure (q). Hence:

Let p be the probability of success, and
Let q be the probability of failure.
Since the experiment succeeds twice as often as it fails, we can write:

p = 2q

Also, since the total probability must sum to 1:

p + q = 1

Substituting p = 2q into this equation:

2q + q = 1 3q = 1 q = 1/3

Therefore, the probability of success is:

p = 2q = 2/3

Step 2: Binomial Distribution
The binomial distribution is given by the formula:

P(X = k) = (nCk) * p^k * q^(n-k)

Where:

P(X = k) is the probability of exactly k successes,
nCk is the number of ways to choose k successes out of n trials (the binomial coefficient),
p is the probability of success on a single trial,
q is the probability of failure on a single trial,
n is the number of trials,
k is the number of successes.
Step 3: Problem Breakdown
We need to find the probability of having at least 4 successes in 6 trials. This means we need to find:

P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)

We are given that n = 6, p = 2/3, and q = 1/3.

Step 4: Calculate Individual Probabilities
We will now calculate the individual probabilities using the binomial distribution formula for k = 4, 5, and 6.

Probability of exactly 4 successes (P(X = 4)):
P(X = 4) = (6C4) * (2/3)^4 * (1/3)^2 = (15) * (16/81) * (1/9) = 15 * (16/729) = 240/729

Probability of exactly 5 successes (P(X = 5)):
P(X = 5) = (6C5) * (2/3)^5 * (1/3)^1 = (6) * (32/243) * (1/3) = 6 * (32/729) = 192/729

Probability of exactly 6 successes (P(X = 6)):
P(X = 6) = (6C6) * (2/3)^6 * (1/3)^0 = (1) * (64/729) * 1 = 64/729

Step 5: Add Probabilities
Now, we add the probabilities to find P(X ≥ 4):

P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6) = (240/729) + (192/729) + (64/729) = 496/729

Thus, the probability of having at least 4 successes in 6 trials is:

P(X ≥ 4) = 496/729

This is the required probability.