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 problem involves a binomial distribution for which the probability \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \) is used, where \( n \) is the number of trials, \( k \) is the number of successes, and \( p \) is the probability of success. Here, \( n = 4 \). Given 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: \[ 6p^2 (1-p)^2 = 4p^3 (1-p) \] Simplify by eliminating common terms: \[ 6(1-p) = 4p \] Rearrange to find \( p \): \[ 6 - 6p = 4p \Rightarrow 6 = 10p \Rightarrow p = \frac{3}{5} \] The probability of getting at least one success is calculated as: \[ P(\text{at least one success}) = 1 - P(\text{no success}) \] Using \( P(\text{no success}) = (1-p)^n \): \[ P(X \geq 1) = 1 - (1-\frac{3}{5})^4 = 1 - (\frac{2}{5})^4 = 1 - \frac{16}{625} = \frac{609}{625} \] Thus, the probability of getting at least one success is \(\frac{609}{625}\).

Answer 2

Let \( p \) and \( q = 1 - p \) be the probabilities of success and failure, respectively. The probability of getting \( r \) successes in \( n \) trials is:

\[ P(r) = \binom{n}{r} p^r q^{n-r}. \]

For \( n = 4 \), the probabilities of two and three successes are equal:

\[ \binom{4}{2} p^2 q^2 = \binom{4}{3} p^3 q. \]

Simplify the binomial coefficients:

\[ 6p^2 q^2 = 4p^3 q. \]

Divide through by \( p^2 q \) (since \( p, q > 0 \)):

\[ 6q = 4p \implies 3q = 2p \implies q = \frac{2}{5}, \; p = \frac{3}{5}. \]

The probability of getting at least one success is:

\[ P(\text{at least one success}) = 1 - P(0), \]

where:

\[ P(0) = \binom{4}{0} p^0 q^4 = q^4 = \left( \frac{2}{5} \right)^4 = \frac{16}{625}. \]

Thus:

\[ P(\text{at least one success}) = 1 - \frac{16}{625} = \frac{609}{625}. \]