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Probability 18 Online
OpenStudy (anonymous):

Twenty marbles are used to spell a word—9 blue ones, 5 red ones, 3 yellow ones and 3 green ones.? Twenty marbles are used to spell a word—9 blue ones, 5 red ones, 3 yellow ones and 3 green ones.? If two marbles are drawn from the set of twenty marbles at random in succession and without replacement, what is the probability (as a reduced fraction) of choosing a marble other than green each time

OpenStudy (kropot72):

On the first draw there are 17 marbles that are not green. So the probability of choosing a marble other than green is 17/20. Having drawn a marble other than green on the first draw, on the second draw there are 16 marbles that are not green. So the probability of choosing a marble other than green is 16/9. The required probability is the product of the above 2 values of probability; \[P(no\ green\ after\ 2\ draws)=\frac{17}{20}\times \frac{16}{19}=you\ can\ calculate\]

OpenStudy (anonymous):

that what I got but their is no answer like that. The answer choice is 1/3, 68/95, 1/11, 3/17

OpenStudy (anonymous):

so I don't know what to do cause I got 16/19 but is not their

OpenStudy (kropot72):

16/19 is not the correct result. When you reduce the fraction by dividing 16 by 4 and dividing 20 by 4 you can get the correct result as follows: \[\frac{17}{20}\times\frac{16}{19}=\frac{17\times4}{5\times19}=you\ can\ calculate\]

OpenStudy (anonymous):

so is 68 over 95

OpenStudy (kropot72):

Yes. That is the correct choice.

OpenStudy (anonymous):

thank so much :)

OpenStudy (kropot72):

You're welcome :)

OpenStudy (anonymous):

is that the reduce franction?

OpenStudy (kropot72):

Yes, it is. the original fraction is \[\frac{17\times16}{20\times19}=\frac{272}{380}\] 4 is a common factor to numerator and denominator. Therefore we can divide the numerator and the denominator by 4 giving:\[\frac{\frac{272}{4}}{\frac{380}{4}}=\frac{68}{95}\] In the above way the fraction has been reduced.

OpenStudy (anonymous):

thank you so much, you save my life :)

OpenStudy (kropot72):

You're welcome :)

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