MHT CET · Maths · Probability
For a Binomial variate \(x\), mean is 2 and variance is 1 , Then odds in favor of \(X=0\) are
- A 4:1
- B 15:1
- C 1:15
- D 1:4
Answer & Solution
Correct Answer
(C) 1:15
Step-by-step Solution
Detailed explanation
Mean \(=n p=2\), variance \(=n p q=1\)
\(\Rightarrow q=\frac{1}{2}, p=\frac{1}{2}\) and \(n=4\)
Now \(P(x=0)={ }^4 C_0 P^0 \cdot q^4=1 \times 1 \times\left(\frac{1}{2}\right)^4=\frac{1}{16}\)
\(\Rightarrow\) odds in favour of \(p(x=0)=\frac{\frac{1}{16}}{1-\frac{1}{16}}=1: 15\)
\(\Rightarrow q=\frac{1}{2}, p=\frac{1}{2}\) and \(n=4\)
Now \(P(x=0)={ }^4 C_0 P^0 \cdot q^4=1 \times 1 \times\left(\frac{1}{2}\right)^4=\frac{1}{16}\)
\(\Rightarrow\) odds in favour of \(p(x=0)=\frac{\frac{1}{16}}{1-\frac{1}{16}}=1: 15\)
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