Hello everyone, I did this exercise:

I have proceeded in this way:

The probability space is initially 7+3=10, So I can compute P(white balls)=0.7 and P(black balls)=0.3.

Then there is the drawing of two balls, So I have computed, through the binomial formula, the probability that the number of white drawn balls was 2,1 or 0:

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So I have obtained P(x=0)=0.09, P(x=1)=0.42, P(x=2)=0.49

Since then, I've calculated the probability to draw a white or a black ball after the drawing of the precedent 2 balls without replacement:

The probability space becomes 10-2=8, so:

If x=0 (two black ball extracted), the probability to draw a white ball is P(white)=7/8=0.875, and P(black)=0.125

If x=1, P(white)=6/8=0.75, P(black)=0.25;

If x=2, P(white)=5/8=0.625, P(black)=0.375

The, in order to write the joint distribution, I compute: The probability that the first two balls are both black and the third is also black->P(x=0 AND y=0), the probability that the first two balls are black and the third is white->P(x=0 AND y=1) ; The probability that the first two balls are one white and one black and the third is black P(x=1 AND y=0), the probability that the first two balls are one white and one black and the third is white P(x=1 AND y=1), the probability that the first two balls are both white and the third is black P(x=2 AND y=0), the probability that the first two balls are both white and the third is white P(x=2 AND y=1)

To do this, I simply compute the products: P(x=0 AND y=0)=P(x=0)*P(black)=0.09*0.125=0.01125

So I compute the Covariance with the formula Cov(X,Y)=

μx= 0.09*0 + 0.42*1 + 0.49*2 = 1.4

μy= 0.3*0 + 0.7*0 = 0.7

E(XY)= (0*0*0.01125) + (0*1*0.105) + (0*2*0.18375) + (1*0*0.07875) + (1*1*0.315) + (2*1*0.30625) = 0.9275

So Covariance ->

V(X)=0.42

So

With this formula, the correlation coefficient is

But my exercise sheet results are:

I have moreover calculated E(X+Y) and I have obtained the same results of my exercise sheet (2.1)

Do you think I did something wrong about covariance and correlation calculation? I know that my results are very similar but I'm not finding any kind of approximation that brings me the results identical to those provided by my professor, so I have a sense that I did something wrong.

I have proceeded in this way:

The probability space is initially 7+3=10, So I can compute P(white balls)=0.7 and P(black balls)=0.3.

Then there is the drawing of two balls, So I have computed, through the binomial formula, the probability that the number of white drawn balls was 2,1 or 0:

S

So I have obtained P(x=0)=0.09, P(x=1)=0.42, P(x=2)=0.49

Since then, I've calculated the probability to draw a white or a black ball after the drawing of the precedent 2 balls without replacement:

The probability space becomes 10-2=8, so:

If x=0 (two black ball extracted), the probability to draw a white ball is P(white)=7/8=0.875, and P(black)=0.125

If x=1, P(white)=6/8=0.75, P(black)=0.25;

If x=2, P(white)=5/8=0.625, P(black)=0.375

The, in order to write the joint distribution, I compute: The probability that the first two balls are both black and the third is also black->P(x=0 AND y=0), the probability that the first two balls are black and the third is white->P(x=0 AND y=1) ; The probability that the first two balls are one white and one black and the third is black P(x=1 AND y=0), the probability that the first two balls are one white and one black and the third is white P(x=1 AND y=1), the probability that the first two balls are both white and the third is black P(x=2 AND y=0), the probability that the first two balls are both white and the third is white P(x=2 AND y=1)

To do this, I simply compute the products: P(x=0 AND y=0)=P(x=0)*P(black)=0.09*0.125=0.01125

**and so on**, until I get this table, that represents the**JOINT DISTRIBUTION**:So I compute the Covariance with the formula Cov(X,Y)=

**E(XY)- μxμy**μx= 0.09*0 + 0.42*1 + 0.49*2 = 1.4

μy= 0.3*0 + 0.7*0 = 0.7

E(XY)= (0*0*0.01125) + (0*1*0.105) + (0*2*0.18375) + (1*0*0.07875) + (1*1*0.315) + (2*1*0.30625) = 0.9275

So Covariance ->

**Cov(X,Y) = 0.9275 - (1.4*0.7) = -0.0525**

Variance of X -:Variance of X -:

V(X)=0.42

So

**V(X)=0.42**and, with the same formula,**V(Y)=0.37****Correlation coefficient:**With this formula, the correlation coefficient is

**ρ(X,Y)= -0.1332**But my exercise sheet results are:

**5b) covariance ≈−0.0467; correlation ≈−0.1667.**I have moreover calculated E(X+Y) and I have obtained the same results of my exercise sheet (2.1)

Do you think I did something wrong about covariance and correlation calculation? I know that my results are very similar but I'm not finding any kind of approximation that brings me the results identical to those provided by my professor, so I have a sense that I did something wrong.

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