TS EAMCET · Maths · Probability
The probability that a mechanic making an error while using a machine on the \(n\)th day is given by \(P\left(E_n\right)=\frac{1}{2^n}\). If he has operated the machine for 4 days, the probability that he had not made a mistake on 3 of 4 days is
- A \(\frac{1}{2}\)
- B \(\frac{1}{4}\)
- C \(\frac{243}{512}\)
- D \(\frac{343}{1024}\)
Answer & Solution
Correct Answer
(C) \(\frac{243}{512}\)
Step-by-step Solution
Detailed explanation
The probability of a mechanic making an error while using a machine on \(n\)th day is \(P\left(E_n\right)=\frac{1}{2^n}\) Machine operated for 4 days, so probability of making an error for 1 st day, 2nd, 3rd and 4th day is \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\)…
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