Probability Calculation: Success vs Failure Rate

This problem involves calculating the probability of a specific outcome in a series of independent trials, which is a classic application of the binomial distribution.

Determining Probabilities of Success and Failure

The problem states that an experiment succeeds twice as often as it fails. Let:

• $p$ be the probability of success in a single trial.
• $q$ be the probability of failure in a single trial.

From the problem statement, we have the relationship: $p = 2q$.

Since the sum of probabilities of success and failure must equal 1, we have:

$$p + q = 1$$

Substitute $p = 2q$ into the equation:

$$2q + q = 1$$

$$3q = 1$$

$$q = \dfrac{1}{3}$$

Now, find $p$:

$$p = 2q = 2 \times \dfrac{1}{3} = \dfrac{2}{3}$$

So, the probability of success ($p$) is $\dfrac{2}{3}$, and the probability of failure ($q$) is $\dfrac{1}{3}$.

Binomial Distribution Parameters

We are considering the next six trials, so the number of trials ($n$) is 6.

We need to find the probability of having "at least four successes". This means the number of successes ($k$) can be 4, 5, or 6.

The binomial probability formula is:

$$P(X=k) = \binom{n}{k} p^k q^{n-k}$$

Where:

• $n$ = number of trials
• $k$ = number of successes
• $p$ = probability of success
• $q$ = probability of failure
• $\binom{n}{k}$ = the binomial coefficient, calculated as $\dfrac{n!}{k!(n-k)!}$

Calculating Probabilities for k=4, k=5, and k=6

Probability of Exactly 4 Successes

For $k=4$:

$$P(X=4) = \binom{6}{4} \left(\dfrac{2}{3}\right)^4 \left(\dfrac{1}{3}\right)^{6-4}$$

Calculate the binomial coefficient $\binom{6}{4}$:

$$\binom{6}{4} = \dfrac{6!}{4!(6-4)!} = \dfrac{6!}{4!2!} = \dfrac{6 \times 5}{2 \times 1} = 15$$

Now, calculate the probability:

$$P(X=4) = 15 \times \left(\dfrac{2^4}{3^4}\right) \times \left(\dfrac{1^2}{3^2}\right) = 15 \times \dfrac{16}{81} \times \dfrac{1}{9} = \dfrac{15 \times 16}{729} = \dfrac{240}{729}$$

Probability of Exactly 5 Successes

For $k=5$:

$$P(X=5) = \binom{6}{5} \left(\dfrac{2}{3}\right)^5 \left(\dfrac{1}{3}\right)^{6-5}$$

Calculate the binomial coefficient $\binom{6}{5}$:

$$\binom{6}{5} = \dfrac{6!}{5!(6-5)!} = \dfrac{6!}{5!1!} = 6$$

Now, calculate the probability:

$$P(X=5) = 6 \times \left(\dfrac{2^5}{3^5}\right) \times \left(\dfrac{1^1}{3^1}\right) = 6 \times \dfrac{32}{243} \times \dfrac{1}{3} = \dfrac{6 \times 32}{729} = \dfrac{192}{729}$$

Probability of Exactly 6 Successes

For $k=6$:

$$P(X=6) = \binom{6}{6} \left(\dfrac{2}{3}\right)^6 \left(\dfrac{1}{3}\right)^{6-6}$$

Calculate the binomial coefficient $\binom{6}{6}$:

$$\binom{6}{6} = \dfrac{6!}{6!(6-6)!} = \dfrac{6!}{6!0!} = 1$$

Now, calculate the probability:

$$P(X=6) = 1 \times \left(\dfrac{2^6}{3^6}\right) \times \left(\dfrac{1^0}{3^0}\right) = 1 \times \dfrac{64}{729} \times 1 = \dfrac{64}{729}$$

Total Probability of At Least Four Successes

To find the probability of at least four successes, we sum the probabilities of exactly 4, 5, or 6 successes:

$$P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$$

$$P(X \ge 4) = \dfrac{240}{729} + \dfrac{192}{729} + \dfrac{64}{729}$$

$$P(X \ge 4) = \dfrac{240 + 192 + 64}{729} = \dfrac{496}{729}$$

Therefore, the probability that in the next six trials there will be at least four successes is $\dfrac{496}{729}$.