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BITSAT 2015 Paper

Section: Mathematics

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Question : 123 of 150

Marks: +1, -0

If in a binomial distribution n = 4, P(X = 0) =

\frac{16}{81}

, then P(X = 4) equals

$\frac{1}{16}$
$\frac{1}{81}$
$\frac{1}{27}$
$\frac{1}{8}$

Solution:

Given n = 4 and P(X = 0) =

\frac{16}{81}

Let p be the probability of success and q that of failure in a trial.

Then, P(X = 0) =

\,\binom{4}{0} p^0 q^4

=

\frac{16}{81}

⇒

q^4

=

\left(\frac{2}{3}\right)^4

(Since

\,\binom{n}{0}

= 1)

⇒ q =

\frac{2}{3}

⇒ p =

\frac{1}{3}

∴ P (X = 4) =

\,\binom{4}{4} p^4 q^0

=

p^4

=

\left(\frac{1}{3}\right)^4

=

\frac{1}{81}

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