you toss 3 coins. what is the probability that you get exactly two heads , given that you at least get one head?
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OpenStudy (anonymous):
wouldnt it be 2/3?
OpenStudy (anonymous):
yes it is..i got it after. thank you
OpenStudy (anonymous):
your welcome!
OpenStudy (zarkon):
why 2/3?
OpenStudy (anonymous):
because there is 3 coins and only 2 possibilities?
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OpenStudy (zarkon):
no
OpenStudy (zarkon):
the answer should be 3/7
OpenStudy (anonymous):
ohhh
OpenStudy (zarkon):
P(2H|at least one head)
=P(2H,at least one head)/P(at least one head)
=P(2H)/[1-P(3 tails)]=(3/8)/(1-1/8)=(3/8)/(7/8)=3/7
OpenStudy (anonymous):
too confusing for me
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OpenStudy (zarkon):
use the definition of conditional probability
\[P(A|B)=\frac{P(A,B)}{P(B)}\]
OpenStudy (zarkon):
list out all the possibilities
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT
only these have at least one head (there are 7 of them)
HHH
HHT
HTH
THH
HTT
THT
TTH
of these only 3 of them have 2 heads
OpenStudy (anonymous):
ok....too confusing for me, but i kinda get it a little.