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Mathematics 13 Online
OpenStudy (anonymous):

A room contains 4 married couples. 3 people are randomly selected. The probability that there is a married couple among the selected people is.......(Write your answer in the form 0.xxx, rounded up to the third digit.)

OpenStudy (anonymous):

4c1*6c1/(8c3) = 0.429 .. check the answer as I have calculated in mind.. but the method is correct

OpenStudy (anonymous):

got the same thing, can u stay on got a couple mroe questions to go

OpenStudy (anonymous):

shoot!!!!

OpenStudy (anonymous):

but be quick.. have a n extra class today .. :(

OpenStudy (anonymous):

if i was to do this by using probability, i.e 1/8 * 1/8 *1/8 or something what would be the way

OpenStudy (anonymous):

this is through probability right?? .. after all probability is Number of arrangement with restriction/All arrangements possible --> This is only formula that you have in probability... everything else is derived from this ... thats the beauty of the subject...

OpenStudy (anonymous):

ok got it

OpenStudy (anonymous):

okk lets keep it this way ... 1st probability that you would find 1 person (he may or may not be a couple) -- > 1 now we have two cases for the second selection --> second person selected is the other partner of the 1st person --> if it is then we need to have the third person anyone if it is not then we need to have the third person as the other partner of either of the two .. =(1/7 + 2/7)

OpenStudy (anonymous):

hope this is clear?

OpenStudy (anonymous):

@shaan can u upload a photo of the answer to d first quest?

OpenStudy (anonymous):

with all steps

OpenStudy (anonymous):

with all steps

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