The probability that a mechanic makes an erro | vedclass

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(C) Let $E_n$ be the event that the mechanic makes an error on the $n$th day. The probability is $P(E_n) = \frac{1}{2^n}$.Let $E_n^c$ be the event tha

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$\frac{243}{512}$

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(C) Let $E_n$ be the event that the mechanic makes an error on the $n$th day. The probability is $P(E_n) = \frac{1}{2^n}$.Let $E_n^c$ be the event that the mechanic does not make an error on the $n$th day. Then $P(E_n^c) = 1 - \frac{1}{2^n}$.For $n = 1, 2, 3, 4$, the probabilities of making an error are $P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{4}, P(E_3) = \frac{1}{8}, P(E_4) = \frac{1}{16}$.The probabilities of not making an error are $P(E_1^c) = \frac{1}{2}, P(E_2^c) = \frac{3}{4}, P(E_3^c) = \frac{7}{8}, P(E_4^c) = \frac{15}{16}$.We want the probability of not making a mistake on exactly $3$ out of $4$ days. This can happen in $4$ mutually exclusive ways:$1$. Error on day $1$ only: $P(E_1)P(E_2^c)P(E_3^c)P(E_4^c) = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} = \frac{315}{1024}$$2$. Error on day $2$ only: $P(E_1^c)P(E_2)P(E_3^c)P(E_4^c) = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} = \frac{105}{1024}$$3$. Error on day $3$ only: $P(E_1^c)P(E_2^c)P(E_3)P(E_4^c) = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{8} \cdot \frac{15}{16} = \frac{45}{1024}$$4$. Error on day $4$ only: $P(E_1^c)P(E_2^c)P(E_3^c)P(E_4) = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{1}{16} = \frac{21}{1024}$Summing these probabilities: $\frac{315 + 105 + 45 + 21}{1024} = \frac{486}{1024} = \frac{243}{512}$.

The probability that a mechanic makes an error while using a machine on the $n$th day is given by $P(E_n) = \frac{1}{2^n}$. If he has operated the machine for $4$ days, the probability that he has not made a mistake on $3$ of the $4$ days is:

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