# Problem:
# I toss a coin 10 times. Determine the probability at least two `heads` appear if at least one `heads` appears.

# Answer:

# Let's define:
#   P(A) := the probability that at least two `heads` appear
#   P(B) := the probability that at least one `heads` appears
# And we want to find P(A|B)

# We can think of P(A) in terms of the complement:
#   P(~A) := the probability that **at most** one `heads` appears
# where
#   P(A) = 1 - P(~A)

# Similarly for P(B):
#   P(~B) := the probability that **no** `heads` appear
# where
#   P(B) = 1 - P(~B)

# P(~A) and P(~B) are found using the Binomial distribution for 1 and 0 successes respectively.

N <- 10   # Number of coin tosses
p <- 0.5  # P(Heads) == P(Tails)

prob_not_A <- pbinom(1, size = N, prob = p)
prob_A <- 1 - prob_not_A

prob_not_B <- pbinom(0, size = N, prob = p)
prob_B <- 1 - prob_not_B

# Using Bayes' theorem we have that:
#   P(A|B) * P(B) = P(B|A) * P(A)
# But P(B|A) = 1 (can you see why?)
# So our answer is as follows:
#   P(A|B) = P(A) / P(B)

answer <- prob_A / prob_B

print(answer)