The probability of getting either all heads o | vedclass

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(B) In each trial,three coins are tossed. The total number of outcomes is $2^3 = 8$.The outcomes for all heads $(HHH)$ or all tails $(TTT)$ are $2$.So

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$\frac{3}{32}$

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(B) In each trial,three coins are tossed. The total number of outcomes is $2^3 = 8$.The outcomes for all heads $(HHH)$ or all tails $(TTT)$ are $2$.So,the probability of getting all heads or all tails in one trial is $p = \frac{2}{8} = \frac{1}{4}$.The probability of not getting all heads or all tails is $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.We want the event to occur for the second time in the $3^{rd}$ trial. This means in the first $2$ trials,the event must occur exactly once,and in the $3^{rd}$ trial,it must occur.The probability is given by: $P = (	ext{Probability of } 1 	ext{ success in } 2 	ext{ trials}) 	imes (	ext{Probability of success in } 3^{rd} 	ext{ trial})$.$P = \binom{2}{1} 	imes p^1 	imes q^1 	imes p = 2 	imes \frac{1}{4} 	imes \frac{3}{4} 	imes \frac{1}{4} = \frac{6}{64} = \frac{3}{32}$.

The probability of getting either all heads or all tails for exactly the second time in the $3^{rd}$ trial,if in each trial three coins are tossed,is

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