JEE Main 2009 — Probability Question with Solution

Given that, no of trials = n
Probability of success (p) = $$\frac{1}{4}$$
$$\therefore$$ Probability of no success = 1 - $$\frac{1}{4}$$ = $$\frac{3}{4}$$

As we know, probability of at least one success = 1 - probability of no success

$$\therefore$$ According to the question,

1 - (Probability of no success in n trials) $$\ge$$ $$\frac{9}{10}$$

$$\Rightarrow$$ 1 - P(x = 0) $$\ge$$ $$\frac{9}{10}$$

$$\Rightarrow$$ 1 - $${}^n{C}_0$$$${\left(\frac{1}{4}\right)}^0$$$${\left(\frac{3}{4}\right)}^n$$ $$\ge$$ $$\frac{9}{10}$$

$$\Rightarrow$$ 1 - $${\left(\frac{3}{4}\right)}^n$$ $$\ge$$ $$\frac{9}{10}$$

$$\Rightarrow$$ $${\left(\frac{3}{4}\right)}^n$$ $$\le$$ $$\frac{1}{10}$$

$$\Rightarrow$$ $${\left(\frac{4}{3}\right)}^n$$ $$\ge$$ $$10$$

[Taking log on both sides]

$$\Rightarrow$$ $$n\left({\log }_{10}^4-{\log }_{10}^3\right)$$ $$\ge$$ $${\log }_{10}^{10}$$

$$\Rightarrow$$ $$n\left({\log }_{10}^4-{\log }_{10}^3\right)$$ $$\ge$$ $$1$$

$$\Rightarrow$$ $$n$$ $$\ge$$ $$\frac{1}{\left({\log }_{10}^4-{\log }_{10}^3\right)}$$

$$\therefore$$ Option (D) is correct.