The sum of mean and variance of a Binomial distribution B D → for 3 trials is 2 . 7 , then the binomial distribution B D → is given by

Mean m for B D = n p and variance σ 2 for B D = n p q Given, n p + n p q = 2 . 7 and n = 3 ∴ n p 1 + q = 2 . 7 ⇒ p 1 + q = 2 . 7 3 ⇒ p 1 + q = 0 . 9 ⇒ 1 - q 1 - q = 0 . 9 [ ∵ p + q = 1 ] ⇒ 1 - q 2 = 0 . 9 ⇒ q 2 = 0 . 1 ⇒ q = ± 0 . 3 ∴ q = 0 . 3 [ ∵ q = - 0 . 3 can't possible as 0 ≤ q ≤ 1 ] ⇒ p = 0 . 7 and q = 0 . 3 ∴ B D is given by 0 . 3 + 0 . 7 5

Mathematics - Probability Question with Solution | TestHub

MathematicsProbabilityRandom Variable and its Probability DistributionMedium2 minPYQ_2018

MathematicsMediumsingle choice

The sum of mean and variance of a Binomial distribution $(\vec{B D})$ for $3$ trials is $2.7$ , then the binomial distribution $\vec{B D}$ is given by

Options:

Answer:

Solution:

Mean $(m)$ for $B D = n p$

and variance $(σ^{2})$ for $B D = n p q$

Given, $n p + n p q = 2.7$ and $n = 3$

$∴ n p (1 + q) = 2.7 \Rightarrow p (1 + q) = \frac{2.7}{3}$

$\Rightarrow p (1 + q) = 0.9$

$\Rightarrow (1 - q) (1 - q) = 0.9$ [ $∵$ $p + q = 1$ ]

$\Rightarrow 1 - q^{2} = 0.9$

$\Rightarrow q^{2} = 0.1 \Rightarrow q = \pm 0.3$

$∴ q = 0.3$ [ $∵ q = - 0.3$ can't possible as $0 \leq q \leq 1$ ]

$\Rightarrow p = 0.7$ and $q = 0.3$

$∴ B D$ is given by ${(0.3 + 0.7)}^{5}$

Stream:BITSATSubject:MathematicsTopic:ProbabilitySubtopic:Random Variable and its Probability Distribution

⏱ 2mℹ️ Source: PYQ_2018

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