Probability
Probability
nta_abhyas_2025
Grade None
Question:
NTA Test 14 (Single Choice)
For an admission test of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is $\frac{1}{3}$, then the probability that he is unable to solve less than two problems is
1. \frac{50}{3} \left(\frac{2}{3}\right)^{49}
2. \left(\frac{2}{3}\right)^{50}
3. \frac{50}{3}\left(\frac{1}{3}\right)^{49}
4. \frac{50}{3}\left(\frac{1}{3}\right)^{49}
Step-by-Step Solution
Key Concept: Apply Bayes' theorem to find the posterior probability of a plant given the observation of a non-defective computer.
Using Bayes' theorem, we need to find the probability of a computer coming from plant Tâ given that it is not defective. We have $P(T_1) = \frac{1}{2}$, $P(\overline{D}|T_1) = \frac{20}{21}$, $P(T_2) = \frac{1}{3}$, $P(\overline{D}|T_2) = \frac{8}{9}$, and $P(T_3) = \frac{1}{6}$. First calculate $P(\overline{D}) = P(T_1)P(\overline{D}|T_1) + P(T_2)P(\overline{D}|T_2) + P(T_3)P(\overline{D}|T_3) = \frac{1}{2} \cdot \frac{20}{21} + \frac{1}{3} \cdot \frac{8}{9} + \frac{1}{6} \cdot \frac{7}{8} = \frac{20}{42} + \frac{8}{27} + \frac{7}{48}$. Then $P(T_1|\overline{D}) = \frac{P(T_1)P(\overline{D}|T_1)}{P(\overline{D})} = \frac{12}{23}$. The required answer is $k = 7$.
Correct Answer: 7