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:

$\frac{609}{625}$

The correct answer is Option (1) → $\frac{609}{625}$

In an binomial distribution,

$P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}p^k (1-p)^{n-k}$

$P(X=2)=P(X=3)$  [Given]

$P(X=2)=\begin{pmatrix}4\\2\end{pmatrix} p^2 (1-p)^2=6p^2(1-p)^2$

$P(X=3)=\begin{pmatrix}4\\3\end{pmatrix} p^3 (1-p)=4p^3(1-p)$

$6p^2(1-p)^2=4p^3(1-p)$

$3(1-p)=2p$

$3=5p$

$p=\frac{3}{5}$

$P(X≥1)=1-p(X=0)$

$P(X=0)=\begin{pmatrix}4\\0\end{pmatrix} p^0 (1-p)^4=(1-p)^4$

$=(1-\frac{3}{5})^4$

$=\frac{16}{625}$

$∴1-P(X=0)$

$=1-\frac{16}{625}=\frac{609}{625}$