TS EAMCET · Maths · Probability
In an examination there are four Yes/No type of questions. The probability that the answer by the student to a question without guess to be correct is \(2 / 3\). The probability that a student guesses a correct answer is \(1 / 2\). A student writes the examination either by without guessing answers to all the 4 questions or by guessing answers to all 4 questions. The probability that he attempt the exam by guessing answers to all questions is \(3 / 7\). Given that a student answered at least 3 questions correctly, the probability that he answered all the questions without guessing is
- A \(\frac{13}{15}\)
- B \(\frac{405}{1429}\)
- C \(\frac{1024}{1429}\)
- D \(\frac{2}{15}\)
Answer & Solution
Correct Answer
(C) \(\frac{1024}{1429}\)
Step-by-step Solution
Detailed explanation
Consider the events \(E_1=\) Answer by student without guessing \(E_2=\) Answer by student guessing \(A=\) At least three question correctly \(P\left(E_1\right)=\frac{3}{7}, p\left(E_2\right)=\frac{4}{7}\)…
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