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If the Mean and Variance of a Binomial Distribution Are 4 and 3, Respectively, the Probability of Getting Exactly Six Successes in this Distribution is - Mathematics

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Question

If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is

Options
  • \[^{16}{}{C}_6 \left( \frac{1}{4} \right)^{10} \left( \frac{3}{4} \right)^6\]

     
  • \[^{16}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}\]

  • \[^{12}{}{C}_6 \left( \frac{1}{20} \right) \left( \frac{3}{4} \right)^6\]

  • \[^{12}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^6\]

     

Solution

\[^{16}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}\]

Mean (np) = 4 and Variance (npq) = 3  

\[\therefore q = \frac{3}{4}\]
\[ \Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4}\text{ and } n = 16\]
\[\text{ Let X denotes the number of successes in 16 trials . Then, }  \]
\[P(X = r) = ^{16}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{16 - r} \]
\[ \Rightarrow P(X = 6) = \text{ Probability (getting exactly 6 successes } )\]
\[ = 16 C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10} \]

  Is there an error in this question or solution?
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APPEARS IN

 RD Sharma Solution for Mathematics for Class 12 (Set of 2 Volume) (2018 (Latest))
Chapter 33: Binomial Distribution
MCQ | Q: 20 | Page no. 29
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If the Mean and Variance of a Binomial Distribution Are 4 and 3, Respectively, the Probability of Getting Exactly Six Successes in this Distribution is Concept: Bernoulli Trials and Binomial Distribution.
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