GATE CS 2020 Q.55 (CS Q.45): Probability and Statistics

For $n \ge 2$ , let $a \in \{0, 1\}^n$ be a non-zero vector. Suppose that $x$ is chosen uniformly at random from $\{0, 1\}^n$ . Then, the probability that $\sum_{i=1}^{n} a_i x_i$ is an odd number is _______.

Your answer

Official answer and solution

Official answer: 0.5
From: Official GATE 2020 answer key

Solution

Condition on one coordinate where a_k = 1 to show the inner product modulo 2 is equally likely to be 0 or 1.

The sum $\sum_{i=1}^{n} a_i x_i$ is an integer. It is an odd number if and only if:

\left(\sum_{i=1}^{n} a_i x_i\right) \equiv 1 \pmod 2

Since $a \in \{0, 1\}^n$ is a non-zero vector, there exists at least one index $k \in \{1, 2, \dots, n\}$ such that $a_k = 1$ .

Decompose the sum modulo 2:

S = a_k x_k + \sum_{i \ne k} a_i x_i \equiv x_k + \sum_{i \ne k} a_i x_i \pmod 2

Because $x$ is chosen uniformly at random from $\{0, 1\}^n$ , the coordinate $x_k$ is independent of all other coordinates $\{x_i \mid i \ne k\}$ , with $\Pr[x_k = 0] = \Pr[x_k = 1] = \frac{1}{2}$ .

Conditioning on any fixed assignment of $\{x_i \mid i \ne k\}$ , the partial sum $\sum_{i \ne k} a_i x_i \pmod 2$ evaluates to some constant $c \in \{0, 1\}$ . Then:

\Pr[S \equiv 1 \pmod 2 \mid \{x_i\}_{i \ne k}] = \Pr[x_k \equiv 1 - c \pmod 2] = \frac{1}{2}

By the law of total probability, the unconditional probability is $\frac{1}{2} = 0.5$ .

Final answer

0.5

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2020 Computer Science and Information Technology (CS)
Question: 55 of 65, printed as CS Q.45
Section: Computer Science and Information Technology

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