A fair coin is flipped 10 times and the sequence of heads and tails is noted. In how many outcomes are there in at least 3 tails?a. 904 b. 120 c. 176 d. 24 e. 32

P(X ≥ 3) = 0.94538

Step-by-step explanation:

When flipping a coin, probability of getting a tail is p= 0.5.

Number of times flipped; n = 10

Now, this is a binomial probability distribution problem with the formula;

P(X = x) = C(n, r) × p^(x) × (1 - p)^(n - x)

Probability of getting at least 3 tails is;

P(X ≥ 3) = P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10)

P(3) = C(10, 3) × 0.5³ × (1 - 0.5)^(10 - 3)

P(3) = 0.1172

P(4) = C(10, 4) × 0.5⁴ × (1 - 0.5)^(10 - 4)

P(4) = 0.2051

From online binomial probability calculator, we have the remaining as;

P(5) = 0.2461

P(6) = 0.2051

P(7) = 0.1172

P(8) = 0.0439

P(9) = 0.0098

P(10) = 0.00098

Thus;

P(X ≥ 3) = 0.1172 + 0.2051 + 0.2461 + 0.2051 + 0.1172 + 0.0439 + 0.0098 + 0.00098

P(X ≥ 3) = 0.94538

f(x) = y

f(-3) = -5

step-by-step explanation:

answer: idk

step-by-step explanation: