an unbiased coin is tossed six times. find the probability of the given event. the coin lands heads more than once.

Can you find the probability that is lands heads twice, thrice, four times, etc...

And then add all those probability together?

0.8906

To find the probability of the coin landing heads more than once when tossed six times, we can use the concept of binomial probability.

The probability of getting heads in a single toss of an unbiased coin is 0.5, and the probability of getting tails is also 0.5.

To find the probability of an event happening more than once, we can calculate the probability of the complementary event (not getting heads more than once) and subtract it from 1.

The complementary event here is getting heads either once or not at all.

The probability of getting heads once is:
P(X = 1) = (6 choose 1) * (0.5)^1 * (0.5)^(6-1) = 6 * 0.5 * 0.5^5

The probability of not getting heads at all is:
P(X = 0) = (6 choose 0) * (0.5)^0 * (0.5)^(6-0) = 1 * 1 * 0.5^6

Adding these probabilities together: P(X = 0) + P(X = 1) = 0.5^6 + 6 * 0.5 * 0.5^5

Now, we can find the probability of the given event happening more than once:
P(X > 1) = 1 - (P(X = 0) + P(X = 1))

P(X > 1) = 1 - (0.5^6 + 6 * 0.5 * 0.5^5)

P(X > 1) ≈ 0.65625

Therefore, the probability of the coin landing heads more than once when tossed six times is approximately 0.65625.

To find the probability of the given event, we need to determine the number of favorable outcomes (coin lands heads more than once) and divide it by the total number of possible outcomes.

First, let's calculate the total number of possible outcomes when a coin is tossed six times. Since each coin toss has two possible outcomes (heads or tails), the total number of possible outcomes is 2^6 = 64.

Next, let's determine the number of favorable outcomes. Since we are looking for the coin to land heads more than once, we need to consider the following cases:
1. Two heads: There are 6C2 = 15 ways to choose 2 out of 6 tosses to be heads.
2. Three heads: There are 6C3 = 20 ways to choose 3 out of 6 tosses to be heads.
4. Four heads: There are 6C4 = 15 ways to choose 4 out of 6 tosses to be heads.
5. Five heads: There are 6C5 = 6 ways to choose 5 out of 6 tosses to be heads.
6. Six heads: There is only 1 way to have all 6 tosses land as heads.

Therefore, the number of favorable outcomes is 15 + 20 + 15 + 6 + 1 = 57.

Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes: Probability = Favorable outcomes / Total outcomes = 57/64 ≈ 0.891.

So, the probability of the coin landing heads more than once is approximately 0.891 or 89.1%.