can someone help me with my normal distribution problem please?
Parth (parthkohli):
Magical coins lol
OpenStudy (anonymous):
it's really hard
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OpenStudy (anonymous):
shoot
OpenStudy (anonymous):
probability of head on any one tossis \(\frac{1}{2}\)
probability of getting 6 in a row is \((\frac{1}{2})^6=\frac{1}{2^6}=\frac{1}{64}\)
OpenStudy (anonymous):
satellite, can you help me out please?
OpenStudy (anonymous):
i really need your help
OpenStudy (anonymous):
whats the question?
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Parth (parthkohli):
Mr. Satellite is our expert. Don't underestimate him ;P
OpenStudy (anonymous):
i'll bump it in a few secs
OpenStudy (anonymous):
bumped
OpenStudy (anonymous):
probability of getting atleast 6 heads
= (number of possible ways of getting number of heads greater than or equal to 6) /(total possibilities that can occur in the given process)
= (1)/(2x2x2x2x2x2)
= \[2^{-6}\]