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

Answer: (A)

$\frac{609}{625}$

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:

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