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(A) Let $H$ be the event of getting a head in a coin toss,$P(H) = \frac{1}{2}$. The probability of not getting a head is $P(H') = 1 - \frac{1}{2} = \f

Vedclass · Accepted Answer

$\frac{3}{4}$

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(A) Let $H$ be the event of getting a head in a coin toss,$P(H) = \frac{1}{2}$. The probability of not getting a head is $P(H') = 1 - \frac{1}{2} = \frac{1}{2}$.Let $D$ be the event of getting a $5$ or $6$ in a dice throw,$P(D) = \frac{2}{6} = \frac{1}{3}$. The probability of not getting a $5$ or $6$ is $P(D') = 1 - \frac{1}{3} = \frac{2}{3}$.The man starts with the coin and they alternate. He wins if he gets a head before getting a $5$ or $6$.This can happen in the following ways:$1$. He gets a head on the first toss: $P_1 = \frac{1}{2}$.$2$. He fails on the first toss,fails on the first dice throw,and gets a head on the second toss: $P_2 = (\frac{1}{2} 	imes \frac{2}{3}) 	imes \frac{1}{2}$.$3$. He fails on the first two tosses and first two dice throws,and gets a head on the third toss: $P_3 = (\frac{1}{2} 	imes \frac{2}{3})^2 	imes \frac{1}{2}$.The total probability is the sum of this infinite geometric series:$P = \frac{1}{2} + (\frac{1}{2} \cdot \frac{2}{3}) \cdot \frac{1}{2} + (\frac{1}{2} \cdot \frac{2}{3})^2 \cdot \frac{1}{2} + \dots$$P = \frac{1}{2} [1 + \frac{1}{3} + (\frac{1}{3})^2 + \dots]$Using the sum formula for an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=\frac{1}{3}$:$P = \frac{1}{2} 	imes \frac{1}{1 - 1/3} = \frac{1}{2} 	imes \frac{1}{2/3} = \frac{1}{2} 	imes \frac{3}{2} = \frac{3}{4}$.

$A$ man alternately tosses a coin and throws a dice,beginning with the coin. The probability that he gets a head in the coin before he gets a $5$ or $6$ in the dice is

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