Question

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

Solution

Here n = 6
Let p be the probabilities of getting a head
$therefore space space straight p space equals space 1 half comma space space straight q space equals space 1 minus 1 half space equals space 1 half$
P(at least four heads) =P(4) + +P(5) + P(6)
$equals space straight C presuperscript 6 subscript 4 space open parentheses 1 half close parentheses to the power of 4 space space open parentheses 1 half close parentheses to the power of 4 space plus space straight C presuperscript 6 subscript 5 space open parentheses 1 half close parentheses space open parentheses 1 half close parentheses to the power of 5 plus space straight C presuperscript 6 subscript 6 space open parentheses 1 half close parentheses to the power of 6 space equals space open parentheses 1 half close parentheses to the power of 6 space left square bracket straight C presuperscript 6 subscript 4 space plus space straight C presuperscript 6 subscript 5 space plus space straight C presuperscript 6 subscript 6 right square bracket space equals space 1 over 64 open square brackets fraction numerator 6 cross times 5 over denominator 1 cross times 2 end fraction plus 6 over 1 plus 1 close square brackets space equals space 1 over 64 left parenthesis 15 plus 6 plus 1 right parenthesis space equals space 22 over 34 space equals space 11 over 32$

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Probability

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

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