A program consists of two modules

[TomJerry]

Question:
A program consists of two modules. The number of errors X in the first module and the number of errors in the second module have the joint distribution:

P(0,0) = P(0,1) = P(1,0) = 0.20
P(1,1) = P(1,2) = P(1,3) = 0.10
P(0,2) = P(0,3) = 0.05

Find
i) the marginal distribution of X and Y
ii) the probability of number errors in the first module.
iii) the probability distribution of total number of errors in the program.
iv) if the errors in the two modules occur independently


Solution

i)
h(y) = 0.40 + 0.30 + 0.15 + 0.15 Is this correct

&

g(x) = 0.50 + 0.50 Is this correct

ii)
g(x) = 0.50 + 0.50 Is this correct

iii)
total = 1 Is this correct

iv)
stuck here

 

[mmwave]

Hi there,

Thank you for your question. I would be happy to help you understand and solve this problem.

First, let's define the variables in this problem. X represents the number of errors in the first module and Y represents the number of errors in the second module. The joint distribution shows the probabilities of different combinations of X and Y occurring.

i) To find the marginal distribution of X and Y, we need to sum up the probabilities for each value of X and Y separately. For example, to find the marginal distribution of X, we add the probabilities for X=0 and X=1:

h(y) = P(X=0, Y=y) + P(X=1, Y=y)

= 0.20 + 0.20 + 0.10 + 0.05

= 0.55

Similarly, to find the marginal distribution of Y, we add the probabilities for Y=0, Y=1, Y=2, and Y=3:

g(x) = P(X=x, Y=0) + P(X=x, Y=1) + P(X=x, Y=2) + P(X=x, Y=3)

= 0.20 + 0.10 + 0.05 + 0.15

= 0.50

ii) The probability of number of errors in the first module can be calculated by looking at the marginal distribution of X. We can see that the probability of X=0 is 0.55 and the probability of X=1 is 0.45. Therefore, the probability of number of errors in the first module is 0.55.

iii) To find the probability distribution of total number of errors in the program, we need to consider all possible combinations of X and Y. We can create a table to show the probabilities for each combination:

| X | Y | Probability |

| 0 | 0 | 0.20 |

| 0 | 1 | 0.20 |

| 0 | 2 | 0.05 |

| 0 | 3 | 0.05 |

| 1 | 0 | 0.20 |

| 1 | 1 | 0.10 |

| 1 | 2 | 0.10 |

| 1 | 3 | 0.10 |

Now, to find the probability distribution of total number of errors,