Understanding physical significant of conditional probability in this event

Question

Question: The probability that a randomly chosen record has an error is 0.05. An error in a record is either large or small. Two out of ten errors, on average, are large.

a. What is the probability that a randomly chosen record has a large error

Using Wolfram Mathematica, my sample space is

Since the probability of any record having an error is 0.05 with $\frac{2}{10}$ being a large error, it must be true that

$\frac{2}{10}$ $\cdot$ 0.05= 0.01 is a large error and the probability of a small error is 0.99.

b. Given that a randomly chosen record does not have a large error, what is the probability that it does not have an error.

b. is a conditional probability.

In my attempt, we have

$P\left ( Null Error | Small Error \right )=\frac{P\left ( Null Error \cap Small Error \right )}{P\left ( Small Error \right )} = \frac{0.99}{0.76}$

But this looks odd. It doesn't make sense for me to talk of small or large error in the null error case.

Any help is appreciated.

Answer 1

Since an error is either large or small (and cannot be both at the same time, meaning the events have the null set as intersection), you have $$\mathbb{P}[\text{error}]=\mathbb{P}[\text{small error}] + \mathbb{P}[\text{small error}].$$

As such, $\mathbb{P}[\text{small error}] = 0.04$, and $\mathbb{P}[\text{no error}] = 1 - \mathbb{P}[\text{error}] = 0.99$

Regarding b., not having a large error is not the same as having a small error: $$\{\text{no large error}\} = \{\text{small error}\} \cup \{\text{no error}\},$$ therefore $\mathbb{P}[\text{no large error}] = 0.04 + (1-0.05) = 0.99$, which means $$ \mathbb{P}[\text{no error}\mid\text{no large error}] =\dfrac{\mathbb{P}[\text{no error}]}{\mathbb{P}[\text{no large error}]} = \dfrac{0.95}{0.99} $$