There is one probability question, saying there are three kinds of socks. And after putting those 6 socks into a drawer, socks are pulled two at a time (without seeing socks you pull). What are the odds of pulling a matching pair?
The answer is 1/3. However, I was a little bit confused as I thought the odds could be 1/5. My logic was like that as the left hand pulls one sock, there are 5 socks remaining in the drawer. And the right hand must pulls the other one that matches. Thus the probability, based on what I thought, should be 1/5. Could anyone help me? Thanks a lot.
Best Answer
You are completely correct.
Simultaneously or not is completely irrelevant.
There are $\binom62=15$ possible choices and $3$ of them are matching choices.
That gives probability $\frac3{15}=\frac15$.
Also your logic is correct and in my view even more elegant.
addendum. The correct interpretation of the probability question might have been: draw $3$ times a pair of socks without replacement. What is the probability that at least one of the pairs is matching?
It is handsome to calculate the probability of the complement. As shown above the probability that the first pair is not matching is $\frac45$. Assume that this happens and let's say that from socks $A_1,A_2,B_1,B_2,C_1,C_2$ we have drawn $A_1B_1$. Then $A_2,B_2,C_1,C_2$ are left and no matching pair will appear if the second draw will be one of $4$ equiprobable possibilities: $A_2C_1$, $A_2C_2$, $B_2C_1$ and $B_2C_2$. There are $6$ possibilities in total so there is chance of $\frac46=\frac23$ that this happens.
Then we come out on probability $\frac45\frac23=\frac8{15}$ that no matching pair is drawn. The probability that at least one matching pair has been drawn is $1-\frac8{15}=\frac7{15}$.