Question

Let \(Y\) be the number of successes throughout \(n\) independent repetitions of a random experiment with probability of success \(p=\frac{1}{4}\). Determine the smallest value of \(n\) so that \(P(1 \leq Y) \geq 0.70\)

Short answer

The smallest number of independent repetitions 'n' so that the probability of at least 1 success is greater than or equal to 0.70 is 4.

Step-by-step solution

Use the Complementary Rule

We start by using the formula for the probability of exactly 'k' successes in 'n' repetitions of a binomial experiment. This is given by \(P(Y=k) = C(n, k) * p^k * (1-p)^{n-k}\) where 'n' is the number of experiments, 'k' is the number of successes, 'p' is the probability of success, and 'C(n, k)' is the binomial coefficient. Then we use the Complementary Rule which states that P(A')=1-P(A) to calculate 'P(1≤Y)=1-P(Y=0)'.

Compute P(Y=0)

Now, we substitute 'k' with '0' in the previously mentioned formula, and obtain \(P(Y=0)=C(n, 0) * p^0 * (1-p)^{n-0}= (1-p)^n \). In our case, p=1/4, so \(P(Y=0)=(1-1/4)^n= (3/4)^n \).

Solving for n

By setting the inequality \(1-P(Y=0)=1- (3/4)^n ≥ 0.70\), we can solve for 'n'. It simplifies to \( (3/4)^n ≤ 0.30\). This inequality doesn't have an exact solution because 'n' must be an integer. In such cases, we round up to the next highest integer.

Calculate the smallest 'n' that satisfies the inequality.

Solving the inequality \( (3/4)^n ≤ 0.30\), we find that n is approximately 3.11. But since 'n' must be an integer, the smallest number of independent repetitions is 4.