Suppose that 3 balls are chosen without replacement from an urn consisting of 5 white and 8 red balls. Let Xi equal 1 if the ith ball selected is white and let it equal 0 otherwise. Give the joint probability mass function of X1 , X2.

The joint probability mass function table is provided below.

Step-by-step explanation:

The experiment consist of drawing 3 balls from an urn containing 5 white and 8 red balls.

The random variable Xi is defined as follows:

Xi = 1; if the ith ball is white

Xi = 0; otherwise.

Then X₁ = the 1st ball is white and X₂ = the 2nd ball is white.

The sample space is as follows:

S = {(X₁ = 0, X₂ = 0), (X₁ = 0, X₂ = 1), (X₁ = 1, X₂ = 0), (X₁ = 1, X₂ = 1)}

Compute the probability of each event in the sample space as follows:

The joint probability mass function table is provided below.

Step-by-step explanation:

I don't know what constructions you were taught.

a "similar" triangle is a triangle with exactly the same angles as the other triangle, but the lengths of all sides are stretched or shortened by the same scaling factor f.

by saying 1:2 she means the second triangle should have sides with twice the lengths of the first triangle (f=2).

and the extra challenge - same basic thing. she allows you to pick one of the two triangles as reference. and then you need to draw a third triangle (again with the same angles) with the side lengths extended by the scaling factor f of 4/3.

I would draw the triangles right on top of each other with the same starting corner (let's call it A) for all 3.

we would get the triangles ABC, AMN and AXY.

the points B and C would be then halfway on AM and AN.

and M and N would then a bit before X and Y on AX and AY.

the beauty is, you only need to construct 2 sides of every new triangle down to the new endpoints. the third side is automatically scaled correctly, and you only need to connect these new endpoints.

let's assume you draw a triangle (just very simple) ABC with all side lengths being 3. so, AB=3, AC=3, BC=3.

now you draw AMN by extending AB and AC to a side length of 6 (f=2) creating M and N, and you connect M and N.

and then you can create the third triangle AXY by extending AM and AN by a factor of 4/3 to side lengths of 8 (4/3 × 6 = 8) creating new end points X and Y. and you connect X and Y.

and that is it. all 3 triangles are similar (the same angles), and all sides of a triangle have the same length ratio to the sides of the other triangle(s).