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 head, then $P\left(E/F\right)$ is equal to

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

Answer 1

Answer: (A)

$3/4$

$P\left(E/F\right)=\frac{P(E\bigcap F)}{P\left(F\right)}$

The total chances are 8 because it is a three coin trail.

The probability of the event F of getting head in first trail is independent of the other 2 trails so it becomes;

$P\left(F\right)=\frac{1}{2}$

The probability that the 3 trails have atleast 2 heads when the first trail is a head we can find it by slotting lets see the possibilities :

$HHH$

$HHT$

$HTH$

$P(E\bigcap F)=\frac{3}{8}$

$\frac{P(E\bigcap F)}{P\left(F\right)}=\frac{\frac{3}{8}}{\frac{1}{2}}=\frac{3}{4}$