Question

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 1

Answer: (C)

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 lst day, $2 nd , 3 rd$ and 4 th day is $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}$ respectively.
Now, the probability that it had not make a mistake on 3 out of 4 is
$=\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{5}{16}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{7}{8} \cdot \frac{15}{16}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{8} \cdot \frac{15}{16}$
$+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{1}{16}$
$=\frac{315}{1024}+\frac{105}{1024}+\frac{45}{1024}+\frac{21}{1024}=\frac{486}{1024}=\frac{243}{512}$