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)
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:
- X = 1 and Y = 0, which happens with probability 0.4 * 0.2 = 0.08.
- 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.
Similar Questions
The random variable X has Bernoulli distribution with parameter 𝑝 = 14a) Calculate 𝐸(𝑋) and 𝑉𝑎𝑟(𝑋) (2marks)b) When n independent Bernoulli trials are carried out with 𝑝 = 14, the number of successes,Y, can be modelled by another distribution.i) State the distribution of Y. (1mark)ii) Given that the variance of Y is 4, determine 𝐸(𝑌) (1mark)
Let 𝑋X and 𝑌Y be independent and identically distributed Geometric random variables with parameter 0.80.8.Find 𝑃(𝑋=4∣𝑋+𝑌=8).P(X=4∣X+Y=8).Enter the answer correct to two decimal places.
The joint pdf of two continuous random variables 𝑋X and 𝑌Y is given by𝑓𝑋𝑌(𝑥,𝑦)={4𝑥𝑦1440≤𝑥≤14,0≤𝑦≤140otherwisef XY (x,y)={ 14 4 4xy 0 0≤x≤14,0≤y≤14otherwise Are 𝑋X and 𝑌Y independent?YesNo
The joint distribution of 𝑋X and 𝑌Y is given by 𝑓𝑋𝑌(𝑥,𝑦)=916×4𝑥+𝑦,f XY (x,y)= 16×4 x+y 9 ,where 𝑇𝑋,𝑇𝑌∈{0,1,2,…}.T X ,T Y ∈{0,1,2,…}.1 pointFind the probability mass function of 𝑋+𝑌X+Y.𝑘916⋅4𝑘k 16⋅4 k 9 (𝑘+1)916⋅4𝑘(k+1) 16⋅4 k 9 (𝑘+1)916⋅4𝑘+1(k+1) 16⋅4 k+1 9 𝑘916⋅4𝑘+1k 16⋅4 k+1 9
Let the discrete RV 𝑋~𝑈[−2,2] (Uniform dist.). Let 𝑌 = 𝑋2a) What values X and Y can take? Find pdf’s of both X and Y.b) Compute the joint pdf, 𝑓𝑋𝑌(𝑥𝑖, 𝑦𝑖)c) Compute the E(X) and E(Y)d) Compute the Cov(X,Y)e) Compute the 𝜌𝑋𝑌 = 𝐶𝑜𝑟(𝑋, 𝑌).f) Are X and Y independent? Prove it.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.