Question

A coin is tossed three times in succession. If $E$ is the event that there are at least two heads and $F$ is the event in which first throw is a head, then $P\left(E/F\right)$ is equal to:

• A. $3/4$
• B. $3/8$
• C. $1/2$
• D. $1/8$

Answer 1

The correct option is A $3/4$

$S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$

Event $E$: Probability that there is at least two head

$=$ $\{HHT,HHH,HTH,THH\}=\frac{4}{8}$

Event $F$: Probability that the event first throw is a head

$=\{HHH,HHT,HTT,HHT\}=\frac{4}{8}=\frac{1}{2}$

$P\left(E/F\right)=$$\frac{P(E\cap F)}{P\left(F\right)}$ $=P\left(EventE\right)/P\left(EventB\right)=$ $\frac{2/8}{1/2}=\frac{1}{2}$