A coin is tossed 6 times. What is the probability of getting at least two heads?

Solution

Here n = 6
Let p be the probability of getting a head
$therefore space space space straight p space equals space 1 half comma space space space straight q space equals space 1 minus straight p space equals space 1 minus 1 half space equals space 1 half$
P(at least 2 heads) = 1 - $left square bracket straight P left parenthesis 0 right parenthesis space plus space straight P left parenthesis 1 right parenthesis right square bracket$
$equals space open square brackets 1 minus straight C presuperscript 6 subscript 0 open parentheses 1 half close parentheses to the power of 6 plus space straight C presuperscript 6 subscript 1 space open parentheses 1 half close parentheses to the power of 1 space open parentheses 1 half close parentheses to the power of 5 close square brackets equals space open square brackets 1 minus space cross times open parentheses 1 half close parentheses to the power of 6 space plus space 6 space cross times space open parentheses 1 half close parentheses to the power of 6 close square brackets space equals space 1 minus 1 over 64 left square bracket 1 plus 6 right square bracket space equals 1 minus 7 over 64 space equals space 57 over 64$

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Probability

A coin is tossed 6 times. What is the probability of getting at least two heads?

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