Question:

Step 1: In a binomial distribution with $n=5$ independent trials, the probability of exactly $k$ successes is given by $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ For $n=5$, this formula becomes: $$P(X=k) = \binom{5}{k} p^k (1-p)^{5-k}$$ Step 2: The probabilities of exactly 1 and 2 successes are given as: $$P(X=1) = \binom{5}{1} p^1 (1-p)^{5-1} = 5p(1-p)^4 = 0.4096$$ $$P(X=2) = \binom{5}{2} p^2 (1-p)^{5-2} = 10p^2(1-p)^3 = 0.2048$$ Step 3: To determine the value of $p$, we form the ratio of $P(X=2)$ to $P(X=1)$: $$\frac{P(X=2)}{P(X=1)} = \frac{10p^2(1-p)^3}{5p(1-p)^4}$$ Simplifying the algebraic expression: $$\frac{10p^2(1-p)^3}{5p(1-p)^4} = \frac{2p}{1-p}$$ Now, substitute the given numerical values: $$\frac{0.2048}{0.4096} = \frac{1}{2}$$ Equating the simplified algebraic expression with the numerical ratio, we get: $$\frac{2p}{1-p} = \frac{1}{2}$$ Step 4: Solve the equation for $p$: $$2p \times 2 = 1 \times (1-p)$$ $$4p = 1-p$$ $$5p = 1$$ $$p = \frac{1}{5}$$ Step 5: Calculate the probability of getting exactly 3 successes using the derived value of $p = \frac{1}{5}$: $$P(X=3) = \binom{5}{3} p^3 (1-p)^2$$ Substitute $p = \frac{1}{5}$ and $1-p = 1 - \frac{1}{5} = \frac{4}{5}$: $$P(X=3) = \binom{5}{3} \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^2$$ $$P(X=3) = 10 \times \frac{1^3}{5^3} \times \frac{4^2}{5^2}$$ $$P(X=3) = 10 \times \frac{1}{125} \times \frac{16}{25}$$ $$P(X=3) = \frac{10 \times 1 \times 16}{125 \times 25}$$ $$P(X=3) = \frac{160}{3125}$$ To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor, which is 5: $$P(X=3) = \frac{160 \div 5}{3125 \div 5} = \frac{32}{625}$$