MHT CET · Maths · Probability
A man is know to speak truth 3 out of 4 times. He throws a die and reports that it is 6 . Then the probability that it is actually 6 is
- A \(\frac{3}{4}\)
- B \(\frac{1}{4}\)
- C \(\frac{3}{8}\)
- D \(\frac{5}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{8}\)
Step-by-step Solution
Detailed explanation
Probability of man speaking truth \(=\frac{3}{4} \Rightarrow\) Probability of telling
\(\text { lies }=\frac{1}{4}\)
Probability of a die actually showing \(6=\frac{\left(\frac{1}{6} \times \frac{3}{4}\right)}{\left(\frac{1}{6} \times \frac{3}{4}\right)+\left(\frac{5}{6} \times \frac{1}{4}\right)}=\frac{3}{8}\)
\(\text { lies }=\frac{1}{4}\)
Probability of a die actually showing \(6=\frac{\left(\frac{1}{6} \times \frac{3}{4}\right)}{\left(\frac{1}{6} \times \frac{3}{4}\right)+\left(\frac{5}{6} \times \frac{1}{4}\right)}=\frac{3}{8}\)
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