A coin is biased so that a head is twice as l | vedclass

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(A) Let the probability of getting a tail be $P(T) = p$. Then the probability of getting a head is $P(H) = 2p$.Since $P(H) + P(T) = 1$,we have $2p + p

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$\frac{2}{9}$

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(A) Let the probability of getting a tail be $P(T) = p$. Then the probability of getting a head is $P(H) = 2p$.Since $P(H) + P(T) = 1$,we have $2p + p = 1$,which gives $3p = 1$,so $p = \frac{1}{3}$.Thus,$P(T) = \frac{1}{3}$ and $P(H) = \frac{2}{3}$.We need the probability of getting $2$ tails and $1$ head in $3$ tosses.The number of ways to arrange $2$ tails and $1$ head is given by the binomial coefficient $\binom{3}{1} = 3$ (the arrangements are $TTH, THT, HTT$).The probability for each arrangement is $P(T) 	imes P(T) 	imes P(H) = \left(\frac{1}{3}ight)^2 	imes \frac{2}{3} = \frac{2}{27}$.Therefore,the total probability is $3 	imes \frac{2}{27} = \frac{6}{27} = \frac{2}{9}$.

$A$ coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed $3$ times,then the probability of getting two tails and one head is-

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