José Silva
New member
- Joined
- Mar 31, 2020
- Messages
- 3
A box contains 4 parts, two of them are defective . To identify the defective parts, successive extractions of a part are carried out, without repositioning. Let X and Y be the real random variables that represent, respectively, the number of extractions performed to detect the first defective part and the number of additional extractions required to detect the second defective part.
a)Check that:
P(X=x, Y=y)=1/6, (x,y) ∈ S={(x,y)∈ {1,2,3}×{1,2,3}:x+y≤4}.
b)Calculate the probability of being needed
i)fewer extractions to detect the second defective part than to detect the first;
ii)at least 3 extractions to detect the two defective parts.
c)Show that X and Y follow the same law of probability.
d)Determine Cºv(X,Y).
e)Construct the probability function of X conditioned by Y = 2 and calculate E (X / Y = 2).
Sorry if its poorly translated, the original problem wast in english and i had to translate it.
a)Check that:
P(X=x, Y=y)=1/6, (x,y) ∈ S={(x,y)∈ {1,2,3}×{1,2,3}:x+y≤4}.
b)Calculate the probability of being needed
i)fewer extractions to detect the second defective part than to detect the first;
ii)at least 3 extractions to detect the two defective parts.
c)Show that X and Y follow the same law of probability.
d)Determine Cºv(X,Y).
e)Construct the probability function of X conditioned by Y = 2 and calculate E (X / Y = 2).
Sorry if its poorly translated, the original problem wast in english and i had to translate it.
