Q49 / 110 · VITEEE 2022 · Mathematics

If the mean and variance of a binomial variate \(x\) are respectively \(\frac{35}{6}\) and \(\frac{35}{36}\), then the probability of \(x > 6\) is :

  • A
    \(\frac{1}{6^2}\)
  • Correct answer
    B
    \(\frac{5^7}{6^7}\)
  • C
    \(\frac{1}{7^6}\)
  • D
    \(\frac{1}{6^7}+\frac{1}{6^7}\)

Solution

Given mean \(=\mathrm{np}=\frac{35}{6}\) ...(i) and variance \(=n p q=\frac{35}{36}\) ...(ii) Solving (i) and (ii), we get \(\mathrm{q}=\frac{1}{6}, \mathrm{p}=1-\frac{1}{6}=\frac{5}{6}\) \(\therefore \quad\) we get \(\mathrm{n}=7\) \(\therefore \quad \mathrm{P}(\mathrm{x}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{p}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}}\) for Binomial distribution where \(r=7\) \(\therefore \quad \mathrm{P}(\mathrm{x} > 6)={ }^7 \mathrm{C}_7\left(\frac{5}{6}\right)^7 \cdot\left(\frac{1}{6}\right)^{7-7}=\left(\frac{5}{6}\right)^7\)

Hence, the correct answer is option: 'B'.

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