Two bags contain $10$ coins each, and the coins in each bag are numbered from $1$ to $10$. One coin is drawn at random from each bag. The probability that one coin has the value $1,2,3$ or $4$ and the other coin has the value $7,8,9$ or $10$ is? I think the answer is $4/25$ but, my friend disagrees and says it is $8/25$
Solution: Since the probability of getting one the $4$ four numbers in a bag of $10$ is $4/10$ and the probability for getting the other $4$ coins in a bag of $10$ is also $4/10$, the total probability is $(4/10)\cdot(4/10)$ which is $4/25$. My friend's approach is that we don't know which bag we are going to choose first. We can either choose $1,2,3$ or $4$ from the $1$st or the $2$nd bag thus $2\cdot(4/25)$ which is $8/25$.