Question

An unbalanced coin is weighted so that the probability of heads is 0.55. The coin is tossed ten times. (a) What is the probability of getting exactly 6 heads?

Answer 1

The probability of getting exactly 6 heads when a biased coin is tossed 10 times is approximately 0.2126, or 21.26%.

Step-by-step explanation:

To find the probability of getting exactly 6 heads when a biased coin is tossed 10 times, we can use the binomial probability formula. The formula is:

P(X = k) = C(n, k) * p^k * (1 - p)^(n-k)

Where:

• n is the number of trials (10 in this case)
• k is the number of successful outcomes (6 heads in this case)
• p is the probability of success (0.55 in this case)
• C(n, k) is the number of ways to choose k successes from n trials (10 choose 6, which is 210 in this case)

Using the values in the formula, we can calculate:

P(X = 6) = C(10, 6) * (0.55)^6 * (1 - 0.55)^(10-6)

P(X = 6) = 210 * (0.55)^6 * (0.45)^4

P(X = 6) ≈ 0.2126

Therefore, the probability of getting exactly 6 heads when the coin is tossed 10 times is approximately 0.2126, or 21.26%.