The sum of mean and variance of a Binomial distribution ( → BD ) for 3 trials is 2 . 7 , then the binomial distribution~ → BD is given by

Correct answer: B. (0 . 3 + 0 . 7)5 Mean (m) for BD = np and variance ( σ ^ 2 ) for BD = npq Given, np + npq = 2 . 7 ~and n = 3 ∴ np(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 ∴ BD ~is given by& (0 . 3 + 0 . 7)5

BITSAT 2018

Exam: BITSAT
Year: 2018
Subject: Mathematics
Chapter: Probability

BITSAT 2018 · Mathematics · Probability

The sum of mean and variance of a Binomial distribution ( → BD )...

The sum of mean and variance of a Binomial distribution

\left(\overset{→}{BD}\right)

for

3

trials is

2 . 7

, then the binomial distribution~

\overset{→}{BD}

is given by

Options

A $\left(0 . 2 + 0 . 8\right)5$
B $\left(0 . 3 + 0 . 7\right)5$
C $\left(0 . 4 + 0 . 6\right)5$
DNone of these

Correct answer

B. $\left(0 . 3 + 0 . 7\right)5$

Step-by-step solution

Mean $\left(m\right)$ for $BD = np$

and variance \left({σ}^{2}\right) $for$ BD = npq $Given,$ np + npq = 2 . 7 $~and$ n = 3 $$ ∴ np\left(1 + q\right) = 2 . 7 ⇒ p\left(1 + q\right) = \dfrac{2 . 7}{3} $$ ⇒ p\left(1 + q\right) = 0 . 9 $$ ⇒ \left(1 - q\right)\left(1 - q\right) = 0 . 9 $[$ ∵p + q = 1 $]$ ⇒ 1 - q^{2} = 0 . 9 $$ ⇒ q^{2} = 0 . 1 ⇒ q = \pm 0 . 3 $$ ∴ q = 0 . 3 $~ [$ ∵ q = - 0 . 3 $~can't possible as$ 0 \le q \le 1 $]$ ⇒ p = 0 . 7 $~and&$ q = 0 . 3 $$ ∴ BD $~is given by& \left(0 . 3 + 0 . 7\right)5$

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