Question

In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:

• A. \(\frac{609}{625}\)
• B. \(\frac{16}{625}\)
• C. \(\frac{513}{625}\)
• D. \(\frac{112}{625}\)

Answer 1

The Correct Option is A

The scenario involves a binomial distribution described by \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( n \) is the number of trials, \( k \) is the number of successes, and \( p \) is the probability of success. Given \( n = 4 \), and the condition that the probability of 2 successes equals the probability of 3 successes: \[ \binom{4}{2} p^2 (1-p)^2 = \binom{4}{3} p^3 (1-p)^1 \] Solving this equation yields: \[ 6p^2 (1-p)^2 = 4p^3 (1-p) \] Simplifying by removing common factors: \[ 6(1-p) = 4p \] Rearranging to determine \( p \): \[ 6 - 6p = 4p \Rightarrow 6 = 10p \Rightarrow p = \frac{3}{5} \] The probability of achieving at least one success is computed as: \[ P(\text{at least one success}) = 1 - P(\text{no success}) \] Utilizing \( P(\text{no success}) = (1-p)^n \): \[ P(X \geq 1) = 1 - (1-\frac{3}{5})^4 = 1 - (\frac{2}{5})^4 = 1 - \frac{16}{625} = \frac{609}{625} \] Consequently, the probability of obtaining at least one success is \(\frac{609}{625}\).