Probability of coins in a bag 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 $8/25$. 
 A: Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that 
$$p_1 = \frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is 
$$p_2 = \frac{4}{10}$$
So net probability will be a product of $p_1$ and $p_2$ as they are independent events and both need to happen simultaneously.
$$P_1 = \frac{4}{10}\cdot \frac{4}{10} \frac{16}{100}$$
Now we take the second case that the coin drawn from first bag is $1,2,3,4$ and coin from the second bag is $7,8,9,20$. Since both bags are identical, this gives us the probability same as before
$$P_2 = \frac{4}{10}\cdot \frac{4}{10} \frac{16}{100}$$
Summing up $P_1$ and $P_2$ (as we need to find the union and they are mutually exclusive)
$$P = P_1 +P_2 = \frac{32}{100} = \frac{8}{25}$$
A: Another approach to the answer:
Possible combinations of sample space are {(1,1),(1,2)...(1,10),(2,1)...(10,10)}. Thus n(S) = 10x10 = 100.
Possible combinations of coins (event E) are {(1,7),(1,8),(1,9),(1,10),(2,7)...(4,10),(7,1),(7,2)...(10,4)}. So n(E) = 2x4x4 = 32.
(Multiplied by 2 since both outcomes like 1,7 and 7,1 are possible).
P(E) = n(E)/n(S) = 32/100 = 8/25
