Probability of failure of first unit provided that at least one of the two units has failed

Question

The following question was asked in JEE Main 2021:

An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is $0.9$ and that of the second unit is $0.8$. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is $p$, then $98p$ is equal to:

My attempt:

Let us call the event of the first unit not functioning $A$ and the second unit not functioning $B$. Let the event of failure of the instrument be $F$. Then $P(F) = 1 - 0.8 \times 0.9 = 0.28$.

From Bayes' Theorem:

$$P(A|F) = \frac{P(F|A) P(A)}{P(F)} = \frac{1 \times 0.1}{0.28}$$ $$P(B|F) = \frac{P(F|B) P(B)}{P(F)} = \frac{1 \times 0.2}{0.28}$$ Hence, $$p = P(A|F) \times (1 - P(B|F)) = \frac{10}{98}$$ And the answer should be $98p = 10$. However, the given answer is:

$28$

Where did I go wrong?

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The solution given by many websites online is:

$$p = \frac{P(A) \times (1 - P(B))}{P(F)} = \frac{28}{98}$$

Although this arrives at the correct answer, this method is not really convincing, and my method seems to be the right method to me. However, I can't arrive at what exactly is wrong in this solution (or mine).

Answer 1

While $A$ and $B$ are independent, you have incorrectly assumed that they are independent conditional on $F$, which is not true. Instead, we have

$$p=P(A,B^c|F)=\frac{P(F|A,B^c)P(A,B^c)}{P(F)}=\frac{P(F|A,B^c)P(A)P(B^c)}{P(F)}=\frac{1\times 0.1\times 0.8}{0.28}$$

so $98p=28.$

Answer 2

Both of $P(A|F)$ and $P(B|F)$ are meaningless unless event $F$ actually happens, so you need to multiply them both by $P(F)$, which gives

$$P(A|F)P(F)\times(1-P(B|F)P(F))$$

and then divide by $P(F)$ to give you the conditional probabilty.