MHT CET · Maths · Probability
If the mean and variance of a binomial distribution are 4 and 2 respectively, then probability of getting 2 heads is
- A \(\frac{28}{256}\)
- B \(\frac{37}{256}\)
- C \(\frac{128}{256}\)
- D \(\frac{219}{256}\)
Answer & Solution
Correct Answer
(A) \(\frac{28}{256}\)
Step-by-step Solution
Detailed explanation
We have \(n p=4\) and \(n p q=2\)
\(\therefore \mathrm{q}=\frac{2}{4}=\frac{1}{2} \quad \Rightarrow \mathrm{p}=1-\frac{1}{2}=\frac{1}{2} \text { and}\) \( \mathrm{n}\left(\frac{1}{2}\right)=4 \Rightarrow \mathrm{n}=8\)
\(\therefore\) \(P(x=2)={ }^8 C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6=\frac{8 !}{2 ! 6 !}\left(\frac{1}{2}\right)^8\) \(=\frac{8 \times 7}{2 \times 2^3 \times 2^5}=\frac{7}{64}=\frac{28}{256}\)
\(\therefore \mathrm{q}=\frac{2}{4}=\frac{1}{2} \quad \Rightarrow \mathrm{p}=1-\frac{1}{2}=\frac{1}{2} \text { and}\) \( \mathrm{n}\left(\frac{1}{2}\right)=4 \Rightarrow \mathrm{n}=8\)
\(\therefore\) \(P(x=2)={ }^8 C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6=\frac{8 !}{2 ! 6 !}\left(\frac{1}{2}\right)^8\) \(=\frac{8 \times 7}{2 \times 2^3 \times 2^5}=\frac{7}{64}=\frac{28}{256}\)
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