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  4. 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:

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Q. 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:

JEE MainJEE Main 2021Probability - Part 2

Solution:

$P ( X =1)={ }^{5} C _{1} \cdot p \cdot q ^{4}=0.4096$
$P ( X =2)={ }^{5} C _{2} \cdot p ^{2} \cdot q ^{3}=0.2048$
$\Rightarrow \frac{ q }{2 p }=2$
$\Rightarrow q=4 p$ and $p+q=1$
$\Rightarrow p=\frac{1}{5}$ and $q=\frac{4}{5}$
Now
$P ( X =3)={ }^{5} C _{3} \cdot\left(\frac{1}{5}\right)^{3} \cdot\left(\frac{4}{5}\right)^{2}$
$=\frac{10 \times 16}{125 \times 25}=\frac{32}{625}$