Question

A coin is tossed three times, where
$E$ is the event of atleast two heads, $F$ is the event of at most two heads. Then the value of $P\left(E/F\right)$ is:

• A. $\frac{3}{7}$
• B. $\frac{3}{8}$
• C. $\frac{7}{8}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{3}{7}$

The sample space of the given experiment will be:
$S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$
Here, $E:$ at least two heads
And $F:$ at most two heads
$\Rightarrow E=\{HHH,HHT,HTH,THH\}$ and $F=\{HHT,HTH,THH,HTT,THT,TTH,TTT\}$
$\Rightarrow E\cap F=\{HHT,HTH,THH\}$
So, $P\left(E\right)=\frac{4}{8}=\frac{1}{2},P\left(F\right)=\frac{7}{8},P(E\cap F)=\frac{3}{8}$
Now we know that
By definition of conditional probability,
$P\left(E/F\right)=\frac{P(E\cap F)}{P\left(F\right)}$
$\Rightarrow P\left(E/F\right)=\frac{3/8}{7/8}=\frac{3}{7}$