JEE PYQ: Probability Question 6
Question 6 - 2021 (18 Mar Shift 2)
Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to:
(1) $\frac{32}{625}$
(2) $\frac{80}{243}$
(3) $\frac{40}{243}$
(4) $\frac{128}{625}$
Show Answer
Answer: (1)
Solution
$P(X=1) = {}^5C_1 p q^4 = 0.4096$ and $P(X=2) = {}^5C_2 p^2 q^3 = 0.2048$. $\frac{P(X=2)}{P(X=1)} = \frac{2p}{q} = \frac{0.2048}{0.4096} = \frac{1}{2}$. So $q = 4p$ and $p + q = 1 \Rightarrow p = \frac{1}{5}$, $q = \frac{4}{5}$. $P(X=3) = {}^5C_3 \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^2 = 10 \times \frac{16}{3125} = \frac{32}{625}$.
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