An unbiased coin is tossed 8 times. The proba | vedclass

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(B) The total number of outcomes when a coin is tossed $8$ times is $2^8 = 256$.Let $H$ denote head and $T$ denote tail.We want the number of outcomes

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$\frac{5}{128}$

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(B) The total number of outcomes when a coin is tossed $8$ times is $2^8 = 256$.Let $H$ denote head and $T$ denote tail.We want the number of outcomes where $H$ appears consecutively at least $5$ times.Case $1$: Exactly $5$ consecutive heads.- $HHHHHTTT$ (and shifts: $THHHHHT T, TTHHHHHT, TTTHHHHH$): $4$ ways.- $HHHHHTHT$ (and shifts: $THHHHHTH, TTHHHHHT$): $3$ ways.- $HHHHHTTH$ (and shifts: $THHHHHTT, TTHHHHHT$): $3$ ways.- $THHHHHTH$ is already counted.- $HHHHHTHH$ is not possible as it would be $6$ heads.Using the inclusion-exclusion principle or manual counting for sequences of length $8$ with at least $5$ consecutive $H$:- Sequences with $5$ consecutive $H$: $HHHHHTXX$ ($4$ ways),$THHHHHTX$ ($3$ ways),$XXTHHHHH$ ($4$ ways).- Total favorable outcomes are $10$.Probability $= \frac{10}{256} = \frac{5}{128}$.

An unbiased coin is tossed $8$ times. The probability that head appears consecutively at least $5$ times is

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