The probability that exactly 3 heads appear i | vedclass

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(B) Let $E_1$ be the event of getting $2$ or more heads in the first three tosses. $E_1 = \{HHH, HTH, HHT, THH\}$. $P(E_1) = \frac{4}{8} = \frac{1}{2}

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$\frac{5}{16}$

Vedclass · Answer

(B) Let $E_1$ be the event of getting $2$ or more heads in the first three tosses. $E_1 = \{HHH, HTH, HHT, THH\}$. $P(E_1) = \frac{4}{8} = \frac{1}{2}$. Let $E_2$ be the event of getting exactly $3$ heads in $6$ tosses. We need to find $P(E_2 | E_1) = \frac{P(E_2 \cap E_1)}{P(E_1)}$. $E_2 \cap E_1$ represents the event where the first three tosses have $2$ or more heads and the total number of heads in $6$ tosses is exactly $3$. Case $1$: First $3$ tosses have $2$ heads (e.g.,$HHT, HTH, THH$). If the first $3$ tosses have $2$ heads, the remaining $3$ tosses must have exactly $1$ head to make the total $3$ heads. Number of ways for first $3$ tosses with $2$ heads $= \binom{3}{2} = 3$. Number of ways for last $3$ tosses with $1$ head $= \binom{3}{1} = 3$. Total ways $= 3 	imes 3 = 9$. Case $2$: First $3$ tosses have $3$ heads $(HHH)$. If the first $3$ tosses have $3$ heads, the remaining $3$ tosses must have $0$ heads to make the total $3$ heads. Number of ways for first $3$ tosses with $3$ heads $= \binom{3}{3} = 1$. Number of ways for last $3$ tosses with $0$ heads $= \binom{3}{0} = 1$. Total ways $= 1 	imes 1 = 1$. Total favorable outcomes $n(E_2 \cap E_1) = 9 + 1 = 10$. $P(E_2 \cap E_1) = \frac{10}{2^6} = \frac{10}{64} = \frac{5}{32}$. $P(E_2 | E_1) = \frac{5/32}{1/2} = \frac{5}{16}$.

The probability that exactly $3$ heads appear in six tosses of an unbiased coin, given that the first three tosses resulted in $2$ or more heads is

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