Please explain and show all work Originally, an urn contained 15 balls One ball
ID: 3172773 • Letter: P
Question
Please explain and show all work
Originally, an urn contained 15 balls One ball having the number 2 on them Two balls having the number 4 on them Five balls having the number 6 on them Seven balls having the number 8 on them Then, the urn was modified At least three, but not all, of the balls with the number 8 were changed to balls with the number 6. Let B represent this unspecified number of balls with the number 8 that were changed to 6 An equal amount of balls with the number 2 and balls with the number 4 were added to the urn. For example, I might have added three 2's and three 4's. I might not have added any at all. Let a represent this unspecified number of balls of each value that were added to the urn. From this newly modified urn, a single ball will be drawn. Let the random variable X the value of the ball that will be drawn from the newly modified urn. Answer the following questions, noting that sometimes your answers will be expressions involving the unspecified constants a and B:Explanation / Answer
Solution
Back-up Theory
For any function to be pmf, it must satisfy 3 conditions:
Strictly speaking, condition (2) is automatically satisfied if the other two conditions are satisfied.
Now, to work out the solution,
The given pmf of X, p(x), is
x
2
4
6
8
p(x)
(1 + )/(15 + 2)
(2 + )/(15 + 2)
(5 + ) /(15 + 2)
(7 – ) /(15 + 2)
It is clear from the description of , that = 0, 1, 2, ………
From the description of , 3 6.
Now, we will see if the 3 conditions on pmf can give additional inputs to decide the range of and .
Condition (1) does not, since for all values of and stated above all pmf values of positive.
Condition (3) also does not, since whatever be the values of and , sum of pmf is 1.
By Condition (2), (1 + ), (2 + ), (5 + ) and (7 – ) must all be less than (15 + 2), which is again valid for all values of and stated above.
Hence, the 3 conditions on pmf do not give additional inputs to decide the range of and .
Part (a)
As explained above, permissible values are:
for , 0, 1, 2, 3, …………. and for , it is 3, 4, 5,6. ANSWER
Part (b)
P(X is a multiple of 4) = P(X = 4) + P(X = 8) = (9 + - )/ (15 + 2) ANSWER
Part (c)
Mean of X = E(X) = sum{x.p(x); x = 2, 4, 6, 8}
= {(2 + 2) + (8 + 4) + (30 + 6) + (56 - 8)}/(15 + 2)
= (96 + 6 - 2) / (15 + 2) ANSWER
Part (d)
and are called parameters. ANSWER
[Note: in the given case, X is a discrete variable and hence it is more appropriate to call the probabilty function as pmf. pdf is used for continuous variables
x
2
4
6
8
p(x)
(1 + )/(15 + 2)
(2 + )/(15 + 2)
(5 + ) /(15 + 2)
(7 – ) /(15 + 2)