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If in a Binomial Distribution N = 4, P (X = 0) = 16 81 , Then P (X = 4) Equals (A) 1 16 P (B) 1 81(C) 1 27 (D) 1 8 - CBSE (Commerce) Class 12 - Mathematics

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Question

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

\[\frac{1}{81}\] In the given binomial distribution, = 4 and

\[P(X = 0) = \frac{16}{81} \]
\[\text{ Binomial distribution is given by } \]
\[P(X = 0) =^ {4}{}{C}_0 \ p^0 q^{4 - 0} = q^4 \]
\[\text{ We know that P } (X = 0) = \frac{16}{81} \]
\[ \therefore q^4 = \frac{16}{81}\]
\[ \Rightarrow q^4 = \left( \frac{2}{3} \right)^4 \]
\[ \Rightarrow q = \frac{2}{3}\]
\[ \therefore p = 1 - \frac{2}{3} = \frac{1}{3}\]
\[\text{ Then }  , P(X = 4) = ^{4}{}{C}_4 \ p^4 q^{4 - 4} \]
\[ = \left( \frac{1}{3} \right)^4 \]
\[ = \frac{1}{81}\]

 
 
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Solution If in a Binomial Distribution N = 4, P (X = 0) = 16 81 , Then P (X = 4) Equals (A) 1 16 P (B) 1 81(C) 1 27 (D) 1 8 Concept: Bernoulli Trials and Binomial Distribution.
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