Question

A binary number is made up of $16$ bits. The probability of an incorrect bit appearing is $p$ and the errors in different bits are independent of one another. The probability of forming an incorrect number is:
$\left( 1 \right)\dfrac{p}{{16}}$
$\left( 2 \right){p^{16}}$
$\left( 3 \right){}^{16}{C_1}{p^{16}}$
$\left( 4 \right)1 - {\left( {1 - p} \right)^{16}}$

Answer 1

In order to solve this question, first we will find the probability of no incorrect digits that will be $\left( {1 - p} \right)$. Then, we will do the calculation for all the 16 digits and the obtained probability is ${\left( {1 - p} \right)^{16}}$. After that, we will calculate the probability of forming a correct number and then, we will use that probability to find the probability of forming an incorrect number.

Complete step-by-step solution:
Since, it is given that a binary number is made of $16$ bits and the probability of an incorrect bit is $p$.
Here, we will need to first find the probability that none of the digits are incorrect.
Since, each digit has the probability $p$ for being incorrect. So, each digit has a probability of being correct and that probability is $\left( {1 - p} \right)$.
Since, all the 16 digits are correct and independent. So, the probability of being correct for all $16$ the digits is ${\left( {1 - p} \right)^{16}}$.
Since, we need to form an incorrect number. So, we will find the probability of all $16$ digits being correct for forming a correct number that is ${\left( {1 - p} \right)^{16}}$.
Hence, the probability of forming an incorrect number is $1 - {\left( {1 - p} \right)^{16}}$.

Note: If the probability of occurrence is $p$, the probability of not occurring is $\left( {1 - p} \right)$. When the events are independent, we can calculate the probability of all the events by multiplying the probability of each event.