Probability Distributions

Since there are two possible outcomes (head or tail) and each outcome is equally likely, the probability of getting a head is 1/2 or 0.5.

The probability of getting exactly 3 heads when flipping a fair coin 5 times can be calculated using the binomial distribution. The formula for the binomial distribution is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 5, x = 3, p = 0.5, and q = 0.5. Plugging these values into the formula, we get P(3) = (5C3) * 0.5^3 * 0.5^(5-3) = 0.3125.

The probability of getting at least 2 heads when flipping a fair coin 4 times can be calculated by subtracting the probability of getting 0 or 1 head from 1. The probability of getting 0 heads is (0.5)^4 = 0.0625, and the probability of getting 1 head is 4 * (0.5)^4 * (0.5)^3 = 0.25. Therefore, the probability of getting at least 2 heads is 1 - 0.0625 - 0.25 = 0.875.

The mean of a normal distribution is the center of the distribution, and it is equal to the expected value of the random variable. In this case, the mean is given as 10, so the mean of the normal distribution is 10.

The standard deviation of a normal distribution is the square root of the variance. In this case, the variance is given as 4, so the standard deviation is √4 = 2.

The probability of getting a value between 8 and 12 from a normal distribution with a mean of 10 and a standard deviation of 2 can be calculated using the standard normal distribution (z-distribution). First, we need to convert the values 8 and 12 to z-scores using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. Plugging in the values, we get z = (8 - 10) / 2 = -1 and z = (12 - 10) / 2 = 1. Then, we can use a standard normal distribution table or calculator to find the probability of getting a z-score between -1 and 1. The probability is 0.6826.

The probability of getting exactly 5 successes in 10 independent trials, each with a probability of success of 0.4, can be calculated using the binomial distribution. The formula for the binomial distribution is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 10, x = 5, p = 0.4, and q = 0.6. Plugging these values into the formula, we get P(5) = (10C5) * 0.4^5 * 0.6^5 = 0.2461.

The probability of getting at least 3 successes in 10 independent trials, each with a probability of success of 0.3, can be calculated by subtracting the probability of getting 0, 1, or 2 successes from 1. The probability of getting 0 successes is (0.7)^10 = 0.0282, the probability of getting 1 success is 10 * (0.7)^9 * (0.3)^1 = 0.1323, and the probability of getting 2 successes is 45 * (0.7)^8 * (0.3)^2 = 0.0856. Therefore, the probability of getting at least 3 successes is 1 - 0.0282 - 0.1323 - 0.0856 = 0.7461.

The probability of getting exactly 4 successes in 10 independent trials, each with a probability of success of 0.2, can be calculated using the binomial distribution. The formula for the binomial distribution is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 10, x = 4, p = 0.2, and q = 0.8. Plugging these values into the formula, we get P(4) = (10C4) * 0.2^4 * 0.8^6 = 0.2048.

The probability of getting at most 2 successes in 5 independent trials, each with a probability of success of 0.6, can be calculated by adding the probabilities of getting 0, 1, and 2 successes. The probability of getting 0 successes is (0.4)^5 = 0.01024, the probability of getting 1 success is 5 * (0.4)^4 * (0.6)^1 = 0.12288, and the probability of getting 2 successes is 10 * (0.4)^3 * (0.6)^2 = 0.28224. Therefore, the probability of getting at most 2 successes is 0.01024 + 0.12288 + 0.28224 = 0.8438.

The probability of getting at least 3 successes in 5 independent trials, each with a probability of success of 0.7, can be calculated by subtracting the probability of getting 0, 1, or 2 successes from 1. The probability of getting 0 successes is (0.3)^5 = 0.00243, the probability of getting 1 success is 5 * (0.3)^4 * (0.7)^1 = 0.02187, and the probability of getting 2 successes is 10 * (0.3)^3 * (0.7)^2 = 0.07569. Therefore, the probability of getting at least 3 successes is 1 - 0.00243 - 0.02187 - 0.07569 = 0.9688.

The probability of getting exactly 2 successes in 5 independent trials, each with a probability of success of 0.4, can be calculated using the binomial distribution. The formula for the binomial distribution is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 5, x = 2, p = 0.4, and q = 0.6. Plugging these values into the formula, we get P(2) = (5C2) * 0.4^2 * 0.6^3 = 0.4096.

The probability of getting at most 1 success in 4 independent trials, each with a probability of success of 0.5, can be calculated by adding the probabilities of getting 0 and 1 successes. The probability of getting 0 successes is (0.5)^4 = 0.0625, and the probability of getting 1 success is 4 * (0.5)^3 * (0.5)^1 = 0.25. Therefore, the probability of getting at most 1 success is 0.0625 + 0.25 = 0.6875.

The probability of getting at least 2 successes in 4 independent trials, each with a probability of success of 0.3, can be calculated by subtracting the probability of getting 0 or 1 success from 1. The probability of getting 0 successes is (0.7)^4 = 0.2401, and the probability of getting 1 success is 4 * (0.7)^3 * (0.3)^1 = 0.3430. Therefore, the probability of getting at least 2 successes is 1 - 0.2401 - 0.3430 = 0.6723.