In 3 trials of a binomial distribution, the probability of 2 successes is 9 times the probability of 3…
In 3 trials of a binomial distribution, the probability of 2 successes is 9 times the probability of 3 successes. Then the probability of success in each trial is
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{1}{4}\)
\(\frac{1}{5}\)
Solution
In a 3 trials of a binomial distribution, the probability of 2 successes \(=9 \times\) probability of 3 successes
\(\begin{aligned}
& \Rightarrow \quad{ }^3 C_2 p^2 q=9\left({ }^3 C_3 p^3\right) \quad\{\text {where } p+q=1\} \\
& \Rightarrow \quad 3 p^2 q=9 p^3 \Rightarrow q=3 p \\
& \therefore p=\frac{1}{4}=\text { probability of success in each trial. }
\end{aligned}\)
Hence, option (c) is correct.