+1 vote
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Let combinational function $f(\text{a, b, c, d}) = \text{abc}'+\text{ab}'\text{cd}'$ (where $x'$ means complement of $x$). If all inputs are equally probable, then the probability that the function evaluates to True is:
(A) 5/16
(B) 1/4
(C) 3/16
(D) 1/8
edited | 109 views

Answer = \begin{align*} \frac{3}{16} \\ \end{align*}

\begin{align*} f(\text{a, b, c, d}) = \text{abc}'+\text{ab}'\text{cd}' = \text{abc}'d + \text{abc}'d' + \text{ab}'\text{cd}' \end{align*}

edited

Boolean function can be simplified as,

a( bc' + b'c )( bc' + d' ) and also each of input is independent of each other. Let the boolean functio be consiting of 3 terms Term-1: a Term-2: bc' + b'c Term-3: bc'+d'

Now each term has to be true simultaneously. So probability of term 1 being true is 1/2 which will be used in all cases.

Next if we take 2 cases, Case 1, bc' is true which makes Term-2 also to be true will have probability 1/4( bcoz of 2 input volved) and with that Term-3 automatically becomes true bcoz of bc' being true

So case 1 probability is 1/2*1/4.

Now case-2 b'c is true it has probability 1/4 so this makes Term-2 true only and not term-3 for Term-3 we must have d' also true it has probability 1/2.

So case-2 probability is 1/2*1/4*1/2

Total probability is 1/8 + 1/6 = 3/16.