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Let 𝑋∼Bernoulli(0.4)X∼Bernoulli(0.4) and 𝑌∼Bernoulli(0.8)Y∼Bernoulli(0.8) be independent. Define 𝑍=𝑋+𝑌−𝑋𝑌Z=X+Y−XY, find the distribution of 𝑍Z.Bernoulli(0.88)Bernoulli(0.88)Bernoulli(0.12)Bernoulli(0.12)Bernoulli(0.92)Bernoulli(0.92)Bernoulli(0.08)Bernoulli(0.08)

Question

Let 𝑋∼Bernoulli(0.4)X∼Bernoulli(0.4) and 𝑌∼Bernoulli(0.8)Y∼Bernoulli(0.8) be independent. Define 𝑍=𝑋+𝑌−𝑋𝑌Z=X+Y−XY, find the distribution of 𝑍Z.Bernoulli(0.88)Bernoulli(0.88)Bernoulli(0.12)Bernoulli(0.12)Bernoulli(0.92)Bernoulli(0.92)Bernoulli(0.08)Bernoulli(0.08)

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Solution

First, let's understand the Bernoulli distribution. A random variable X follows a Bernoulli distribution if it takes the value 1 with probability of success p and the value 0 with probability of failure q=1-p.

Given X ~ Bernoulli(0.4) and Y ~ Bernoulli(0.8), we know that X takes the value 1 with probability 0.4 and 0 with probability 0.6. Similarly, Y takes the value 1 with probability 0.8 and 0 with probability 0.2.

Now, let's define Z = X + Y - XY. Since X and Y are independent, we can calculate the probability of Z taking the value 1 (success) or 0 (failure).

Z can take the value 1 in two cases:

  1. X = 1 and Y = 0, which happens with probability 0.4 * 0.2 = 0.08.
  2. X = 0 and Y = 1, which happens with probability 0.6 * 0.8 = 0.48.

Adding these probabilities together, we find that Z takes the value 1 with probability 0.08 + 0.48 = 0.56.

Therefore, Z follows a Bernoulli distribution with parameter p = 0.56. The other options (Bernoulli(0.88), Bernoulli(0.12), Bernoulli(0.92), and Bernoulli(0.08)) are incorrect.

This problem has been solved

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