An unbiased coin is tossed six times. Find the probability of obtaining at least 4 heads

Solution

Here n = 6
p = Probability of head = $1 half comma space space straight q space equals space 1 space minus space straight p space equals space 1 minus 1 half space equals 1 half$
P (at least 4 heads) = P (4) + P (5) + P (6)
$equals space straight C presuperscript 6 subscript 4 space open parentheses 1 half close parentheses squared space open parentheses 1 half close parentheses to the power of 4 space plus space straight C presuperscript 6 subscript 5 space open parentheses 1 half close parentheses space open parentheses 1 half close parentheses to the power of 5 space plus space straight C presuperscript 6 subscript 6 space open parentheses 1 half close parentheses to the power of 6 equals space open parentheses 1 half close parentheses to the power of 6 space open square brackets straight C presuperscript 6 subscript 4 space plus straight C presuperscript 6 subscript 5 space plus space straight C presuperscript 6 subscript 6 space close square brackets space equals space 1 over 64 space open square brackets fraction numerator 6 space cross times 5 space over denominator 1 space cross times 2 end fraction space plus 6 over 1 plus 1 close square brackets space equals 1 over 64 open square brackets 15 plus 6 plus 1 close square brackets space equals 22 over 64 equals space 11 over 32$

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Probability

An unbiased coin is tossed six times. Find the probability of obtaining at least 4 heads

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