# Probability of a person correctly guessing at least one number out of the two number another person chooses

Person A randomly chooses a number from 1 to 5 (inclusive) twice, so A ends up with 2 numbers chosen (can be the same number). Person B also makes a random choice from that list (only 1 number). What's the probability that B's choice match at least one of A's choices?

My interpretation of the question is: what's the probability that a randomly chosen number (call it $$c$$) between 1 and 5 is in the random set $$[a, b],~ a$$ and $$b$$are between 1 and 5.

My attempt of solving this question is the following:
There are 2 cases: 1.$$a$$ and $$b$$ are different numbers; 2. $$a$$ and $$b$$ are the same.\

case 1:
if $$a = b$$, $$\mathbb{P}[c \in \{a, b\}] = \mathbb{P}[c = a] = \frac{1}{5}$$

case 2:
if $$a \not= b$$, $$\mathbb{P}[c \in \{a, b\}] = \mathbb{P}[c = a] + \mathbb{P}[c = b]= \frac{2}{5}$$

The probability of case 1 occurring is $$\frac{1}{5}$$. The calculation is similar to case 1. And the probability of case 2 occurring is $$\frac{4}{5}$$ since its the complement of case 1. Then, the answer is $$\mathbb{P}[\text{case 1}] \cdot\frac{1}{5} + \mathbb{P}[\text{case 2}]\cdot\frac{2}{5} = \frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot\frac{2}{5} = \frac{3}{5}$$

However, the correct answer is $$\frac{9}{25}$$. This is confirmed by a simulation I ran.
What did I do wrong?

• I think I got it. $\mathbb{P}[c\in \{a,b\}] = \mathbb{P}[c = a \lor c = b] = \mathbb{P}[c = a] + \mathbb{P}[c = b] - \mathbb{P}[c = a \land c = b] = \frac{1}{5} + \frac{1}{5} - \frac{1}{25} = \frac{9}{25}$ If you have anything else to add, please feel free to do so. Thanks! Commented Aug 24, 2022 at 4:37
• Alternatively $1-4/5*4/5$, the negation of the probability that $c\neq a$ and $c\neq b$. Commented Aug 24, 2022 at 7:18
• Some nitpicking: 1 The term 'ramdomly choose' is a bit weird expression and it should be interpreted as drawing a random number. There is no choice involved. 2 Also the probabilities should be considered to be uniform (this is how 'randomly pick' is colloquially understood but formally it is incomplete). 3 In addition the draws need to be considered independent. (it states 'can be the same number' but is not so clear about the probability or potential correlation). Commented Aug 24, 2022 at 7:23
• Your reasoning and calculations are all completely correct, except for the very last part. You correctly arrived at P = 1/5 * 1/5 + 4/5 * 2/5, but then you incorrectly simplified that to P = 3/5 for some reason. But instead it simplifies to P = 1/25 + 8/25 = 9/25.
– Stef
Commented Aug 24, 2022 at 8:43

Your reasoning is correct up to the very last line: $$\mathbb{P}[\text{case 1}] \cdot\frac{1}{5} + \mathbb{P}[\text{case 2}]\cdot\frac{2}{5} = \frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot\frac{2}{5}$$
But this is not equal to $$\frac{3}{5}$$.
$$\frac{1}{5} \cdot \frac{1}{5} + \frac{4}{5} \cdot\frac{2}{5} = \frac{1 \cdot 1}{25} + \frac{4 \cdot 2}{25} = \frac{9}{25}$$