Die Probability Question + Basics of Conditional Probability

Question

A die is rolled twice. What is the probability of observing:

a) a four and a three

P (obtaining a four and a three) or P(obtaining a three and a four)

therefore

P(obtaining a four)* P(obtaining a three) + P(obtaining a three) * P(obtaining a four)

• =(1/6*1/6)+(1/6*1/6)
• =1/18

Would you say, Let A = event 'obtaining a three' and Let B = event 'obtaining a four'?

How else could you solve this probability?

Is it correct to solve this in terms of 'conditional probability:'

that is P(obtaining a three, given you rolled a four) and P(obtaining a four, given you rolled a three) ??

Answer 1

There is no reason to solve this question using conditional probability, but since you asked, here is the logic of solving it with that:

P(A and B)=P(A|B)$\cdot$P(B)

P(we got 3 in one die|one die is four)$\cdot$ P(one die is four)=$\frac{1}{6}\cdot \binom{2}{1}\frac{1}{6}=\frac{1}{18}$

Answer 2

P(obtaining a four) x P(obtaining a three) + P(obtaining a three) x P(obtaining a four)

=(1/6*1/6)+(1/6*1/6)

This is correct, because the result of each roll is independent from the other. But $(1/6*1/6)+(1/6*1/6) \neq 2/18$

Edit: I have seen, that you have corrected your result.