Find the prob. of getting one head when two coin tossed, if we know at least one of coins is head

Question

I have two coins. I want to find the probability of getting one head if I peek and tell you that at least one of the coins is definitely a head when two coins are tossed? Please let me know the my answer is whether correct or not?

This is my solution When two coins are tossed, we have 4 outcomes are equally likely: HH, HT, TH, TT.

Let A be the event that getting one head $A=\{H\}$, then $P(A)=1$

Let B be the event that at least one of the coins is definitely a head is $B=\{ HT,TH,HH \}$, then $P(B)=3/4$

Then $A∩B=0$ We want $P(A|B)$ (the probability of A given B).

By a standard formula, $P(A|B)=\frac{P(A∩B)}{P(B)}=0$

Actually, I am confusing about A event. What does it mean of getting one head? In my knowledge, I think that getting one head means we only achieve one head from two coins. But it looks like impossible case because when we toss, we often have 4 outcomes such that HH,HT,TH,TT.

Answer 1

As comment of Zoran Loncarevic, the event A will be A={HT,TH} , Thereby $P(A)=1/2$

$A ∩ B=\{HT,TH\} $, thus $P(A ∩ B)=1/2$

Base on conditional prob., we have $P(A|B)=\frac{P(A∩B)}{P(B)} =\frac{1}{2} \times \frac{4}{3}=\frac{2}{3} $

Answer 2

Simple answer: There are four equally-probable possible two-coin tosses, HH, HT, TH, TT, But we know one coin is heads, so that eliminates the TT possibility, leaving HH, HT, and TH. Two of those possibilities have exactly one head, so the answer is 2/3.