A coin is tossed 10 times. What is the probability that at least 8 heads appear?

so you want 8 heads, 9 heads or 10 heads

prob = C(10,8) (1/2)^10 + C(10,9) (1/2)^10 + C(10,10) (1/2)^10
= (1/2)^10 (45 + 9 + 1)
= 55 (1/2)^10 = 55/1024 or appr .0537

To find the probability of getting at least 8 heads out of 10 coin tosses, we can use the concept of probability and the binomial distribution.

The binomial distribution is a probability distribution that represents the number of successes in a fixed number of independent Bernoulli trials (like flipping a coin). In this case, we want to find the probability of getting at least 8 heads in 10 coin tosses.

Let's break this down into three possibilities:
1. Getting exactly 8 heads
2. Getting exactly 9 heads
3. Getting exactly 10 heads

First, we need to calculate the probability of each individual outcome.

The probability of getting a head (H) in a coin toss is 1/2, and the probability of getting a tail (T) is also 1/2. Since each coin toss is independent, we can multiply the probabilities together.

For example, the probability of getting exactly 8 heads (H) in 10 coin tosses would be:

P(HHHHHHHHT) = (1/2)^8 * (1/2)^2 = 1/256

Similarly, the probabilities for 9 heads and 10 heads would be:

P(HHHHHHHHH) = (1/2)^9 * (1/2)^1 = 1/512
P(HHHHHHHHHH) = (1/2)^10 = 1/1024

Now that we have the probabilities of each specific outcome, we can sum them up to find the probability of getting at least 8 heads.

P(at least 8 heads) = P(8 heads) + P(9 heads) + P(10 heads)
= 1/256 + 1/512 + 1/1024
= 1/256 + 2/256 + 1/256
= 4/256
= 1/64

So, the probability of getting at least 8 heads in 10 coin tosses is 1/64.