Answer: [tex]\dfrac{511}{512}[/tex]
Step-by-step explanation:
We know that the total number of faces on a coin [Tail, Heads] =2
The probability of getting a tail = [tex]\dfrac{1}{2}=\dfrac{1}{2}[/tex]
The probability of getting no tail = [tex]1-\dfrac{1}{2}=\dfrac{1}{2}[/tex]
The "at least once" rule says that the when a coin is tossed n times , then the probability of getting at least one tail is given by :-
[tex]\text{P(Atleast one tail)}=1-(\text{P(No tail)})^n[/tex]
Since , n=9
Then, the probability of getting at least one tail is given by :-
[tex]\text{P(Atleast one tail)}=1-(\text{P(No tail)})^9\\\\=1-(\dfrac{1}{2})^9\\\\=1-\dfrac{1}{512}\\\\=\dfrac{511}{512}[/tex]
The probability is [tex]\dfrac{511}{512}[/tex]
