The following are the probabilty distributions for the randomvariables X and Y.
ID: 2953581 • Letter: T
Question
The following are the probabilty distributions for the randomvariables X and Y. X P(X) Y P(Y) 1 0.5 1 0.4 2 2 3 3 If X any Y are independent and the joint probability P(X=1,Y=2) = 0.05 and P(X=2, Y=3) = 0.15, what is the P(X=3)? a. 0.1 b. 0.2 c. 0.3 d. 0.4 e. 0.5 Please explain your answer to this question and I will rateLifesaver for you. Thanks! The following are the probabilty distributions for the randomvariables X and Y. X P(X) Y P(Y) 1 0.5 1 0.4 2 2 3 3 If X any Y are independent and the joint probability P(X=1,Y=2) = 0.05 and P(X=2, Y=3) = 0.15, what is the P(X=3)? a. 0.1 b. 0.2 c. 0.3 d. 0.4 e. 0.5 Please explain your answer to this question and I will rateLifesaver for you. Thanks!Explanation / Answer
This is a relatively easy problem, but it will take some time. Xand Y are independent, so you can multiply chances. P(X=1, Y=2) = 0.05 and P(X=2, Y=3) = 0.15 are given: P(x1) is known, so it is possible to calculate P(y2), which is myfirst step to the solution: 1) Calculate P(y2) P(x1)*P(y2) = 0.05 and (x1) = 0.5, so: 0.5 *P(y2) = 0.05 --> P(y2) = 0.10 P(y1) and p(y2) are known, so we are going to calculate p(y3), sowe can calculate after this p(x2) out of P(X=2, Y=3) = 0.15 2) CalculateP(y3) P(y1) +P(y2)+P(y3)= 1 just as the probability of throwing1,2,3,4,5, or 6 with a dice = 1 and each individual is 1/6 P(y3) = 1- P(y2)-P(y1)= 1- 0.10- 0.4= 0.5 3) Calculate P(x2) P(x2)*p(y3)= 0.15 P(x2) *0.5 = 0.15 --> P(x2)= 0.30 Now, we know P(x1) and P(x2), so we can calculate P(x3) 4) Calculate P(x3) P(x1)+P(x2)+P(x3)=1 P(x3)= 1- P(x2)-P(x1)= 1- 0.30-0.5=0.2 Good Luck!
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