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Mathematics 6 Online
OpenStudy (anonymous):

Statistics question about probability based on combination formula. Please see attached photo for question. part a, i tried => ( 25C6 * 12C0) / 36C6 part b, i did 1-ans for part a. Please assist

OpenStudy (anonymous):

OpenStudy (perl):

what do you have so far

OpenStudy (perl):

there are 25 out of 37 cells capable of replication, so there is a 25/37 probability of capable of replication

OpenStudy (anonymous):

part a) \[\frac{ 25C6 * 12C0 }{ 37C6 } = 0.0761\]

OpenStudy (perl):

probability of choosing 6 cells capable of replication. Since we have without replacement, it should be (25/37)*(24/36)*(23/35)*(22/34)*(21/33)

OpenStudy (perl):

That is the probability of probability (first cell replicates & second cell replicates & third cell & fourth cell & fifth cell & sixth cell)

OpenStudy (anonymous):

Oh ok, is it similar to an at least question?

OpenStudy (perl):

we are using P( A & B ) = P( A ) * P ( B | A )

OpenStudy (perl):

the at least is a little different

OpenStudy (perl):

it is easier if you approach it using complement method

OpenStudy (perl):

P ( at least one is not capable of replication) = 1 - P ( all are capable of replication)

OpenStudy (anonymous):

Ok, thanks a lot.

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