Getting at least one head when tossing nine fair coins

Question

asked Jun 20, 2019 165k views

1 Answer

Kalyan · Answer 1 · 2019-06-24T17:59:37+0000

Answer:

Step-by-step explanation:

Let A represent "getting at least one head in 9 tosses"

Then Â is "getting no heads in 9 tosses"

P(A) = 1 - P(Â)

Â only occurs when all tosses are tails (TTT TTT TTT), so there is only one option of getting no heads.

There are a total of 2⁹ possible outcomes.

So, P(A) = 1 -
(1)/(2^(9))

=
(2^(9))/(2^(9)) -
(1)/(2^(9))

=
(2^(9)-1)/(2^(9))

=
(511)/(512)