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STA2603 ASSIGNMENT 4 FULL DETAILED SOLUTIONS UNISA SEMESTER 1 AND 2 R250,00
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STA2603 ASSIGNMENT 4 FULL DETAILED SOLUTIONS UNISA SEMESTER 1 AND 2
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STA2603 ASSIGNMENT 4 FULL DETAILED SOLUTIONS UNISA SEMESTER 1 AND 2. A very good mark guaranteed. A really good explanation and will really help to achieve good marks and understand for the exam

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  • July 21, 2021
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STA2603

ASSIGNMENT 4




QUESTION 1



(a)
Here, first we have to calculate the marginal probability mass function pX(x) of X.
Now, X can take only three values 1, 2, 3 with respective probabilities P(X=1) = P(X=1,Y=1) +
P(X=1,Y=2) + P(X=1,Y=3) = 0.1 + 0.2 + 0 = 0.3, P(X=2) = P(X=2,Y=1) + P(X=2,Y=2) +
P(X=2,Y=3) = 0 + 0.1 + 0.1 = 0.2 and P(X=3) = P(X=3,Y=1) + P(X=3,Y=2) + P(X=3,Y=3) = 0.1 +
0.2 + 0.2 = 0.5. So, the probability mass function pX(x) of X will be :
P(X=x) = 0.3, if x=1

0.2, if x=2

0.5, if x=3

0, otherwise.

Here, next we have to calculate the marginal probability mass function pY(y) of Y.
Now, Y can take only three values 1, 2, 3 with respective probabilities P(Y=1) = P(X=1,Y=1) +
P(X=2,Y=1) + P(X=3,Y=1) = 0.1 + 0 + 0.1 = 0.2, P(Y=2) = P(X=1,Y=2) + P(X=2,Y=2) +
P(X=3,Y=2) = 0.2 + 0.1 + 0.2 = 0.5 and P(Y=3) = P(X=1,Y=3) + P(X=2,Y=3) + P(X=3,Y=3) = 0 +
0.1 + 0.2 = 0.3. So, the probability mass function pY(y) of Y will be :
P(Y=y) = 0.2, if y=1

0.5, if y=2

0.3, if y=3

0, otherwise.

(b)

Here, we have to say whether the random variables X and Y are independent.

Now, we know that, two discrete random variables X and Y are said to be independent if PX,Y(x,y)
= PX(x).PY(y) for all possible pairs of (x,y) and said to be dependent if PX,Y(x,y) PX(x).PY(y) for at
least one pair (x,y). Now, as we can notice that, 0.1 = PX,Y(1,1) PX(1).PY(1) = 0.3 0.2 = 0.06,
so we can claim that the random variables X and Y are not independent.


(c) Here, we have to calculate P (Y = 2 | Y 2).

Now, this required probability will be:

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