Probability with OR, that doesn't add?

Question

I would have thought that the answer to this question is $1/10 + 1/10=1/5$ due to the "OR" part of the question.

Why does that not work here?

Answer 1

Let A be the event the first pick is 1, let B be the event the 2nd pick is 2.

We are asked to find $P(A\cup B) = P(A) +P(B) - P(A\cap B) = \frac{1}{10} + \frac{1}{10} - \frac{1}{10}\frac{1}{10} = \frac{19}{100} $

as required.

Answer 2

Because there is a finite probability that the first card is 1 AND the second card is 2, which you otherwise count twice. This probability is $\frac{1}{10}\frac{1}{10}$. So your final answer is $\frac{1}{10}+\frac{1}{10}-\frac{1}{10}\frac{1}{10}=\frac{19}{100}$

Answer 3

Probability of first card is "1" and the second any other but "2":

$$\dfrac{1}{10}\dfrac{9}{10}$$

Probability of first card any but "1" and the second is "2":

$$\dfrac{9}{10}\dfrac{1}{10}$$

Probability of first being "1" and the second "2"

$$\dfrac{1}{10}\dfrac{1}{10}$$

All summed together are: $\dfrac{9}{100} + \dfrac{9}{100} + \dfrac{1}{100} = \dfrac{19}{100}$

Answer 4

The reason comes exactly from the fact that you'd be double counting.

This is the probability rule:

$$ P(A\ or\ B) = P(A) + P(B) - P(A\ and\ B)$$

So you add the two events, and then subtract the probability that they both occur simultaneously, which is, in this case, $\frac{1}{100}$