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 $$8/25$$.

• How did you get these answers? Could you please edit this question to include your work? It would help us see where a mistake is. (by the way, your answer is wrong, probably because you are only counting $P(\text{coin}_1\in\{1,2,3,4\}\cap \text{coin}_2\in\{7,8,9,10\}$, and not the reverse) – John Doe Jan 6 '19 at 17:41

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)

$$P = P_1 +P_2 = \frac{32}{100} = \frac{8}{25}$$

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