Final answer:
The probability of getting less than 3 heads when tossing an unbalanced coin with a probability of heads being 0.55 is 0.05422 or 5.422%.
Step-by-step explanation:
To find the probability of getting less than 3 heads when tossing an unbalanced coin with a probability of heads being 0.55, we need to calculate the probability of getting 0, 1, or 2 heads.
We can use the binomial probability formula: P(X=k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes, p is the probability of success, and (n choose k) is the binomial coefficient.
In this case, n=10, p=0.55, and we want to calculate P(X<3) = P(X=0) + P(X=1) + P(X=2).
Plugging in the values, we get:
- P(X=0) = (10 choose 0) * 0.55^0 * 0.45^10 = 0.00042
- P(X=1) = (10 choose 1) * 0.55^1 * 0.45^9 = 0.00691
- P(X=2) = (10 choose 2) * 0.55^2 * 0.45^8 = 0.04689
Adding these probabilities together, we get P(X<3) = 0.00042 + 0.00691 + 0.04689 = 0.05422.
Therefore, the probability of getting less than 3 heads when tossing the coin 10 times is 0.05422 or 5.422%.