Professor Egghead woke up one morning and started dressing for a day in the office. He reached into his sock drawer and felt that there were four individual socks. He knew that two of those socks were black, one was white, and one was green. He grabbed two at random and put them on before leaving for work.
If one of the socks he was wearing was black, what is the probability that the other one was also black?
Friday, February 8, 2008
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7 comments:
i think it is 1/4 but i am not sure
if counting the second sock only, it should be 1/(4-1) = 1/3
if counting both 2 socks, it should be 2/4 * [1/(4-1)] = 1/6
There are a total of six combinations that the socks can be selected. Let's call the four individual socks B1, B2, W, and G (The black socks get unique names because they are unique socks.) With those names, we can write the six combinations as:
1. B1, B2
2. B1, W
3. B2, W
4. B1, G
5. B2, G
6. W, G
We are told that one of the two socks is black so we don't have to concern ourselves with the last option with one white sock and one green sock--this leaves 5 possible ways that one of the socks could be black. Only one combination, out of the five remaining, provides the condition of both socks being black.
Therefor, IF one of the socks is black, then the probability that both are black is 1/5.
^It would be 1 in 3 actually because it says 'what is the probability that the other one was also black?'
Well, He has a black one on already, but it doesnt say it matters which black one. The only thing that matters is that he has one black sock on already and that there is a possiblity that the other one is black, white or green. This means there are three possible options. Therefore, the probability of his other sock being black is 1/3.
HA.
it would be 1/3.
it would be 1/3
because...
the question says that the 2 socks were picked at once which means that the question can be re-phrased as - what is the probability of picking 2 black socks out of 4(2B, 1W, 1G)?
soln:
no. of possibilities = 4 = B
{(BB),(WG),(BW),(BG)}
- since 2 blacks are assumed identical
no.of desirable outcomes = 1 = A
{(BB)}
probability = A/B
sorry for the mistake in the first line. read it as 1/4 and not 1/3
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