Question

Let X denote the number of times a certain numerical control machine will malfunction: 1, 2, or 3 times on any given day. Let Y denote the number of times a technician is called on an emergency call. Their joint probability distribution is given. find p(x y>2)

Answer 1

Answer:

$P(Y = 3 | X = 2) = 0.2857$

Step-by-step explanation:

Given

The question has missing details, and it is poorly formatted.

See attachment for complete question

From the attachment, we have:

Calculate: $P(Y = 3 | X = 2)$

This is calculated as:

$P(Y = 3 | X = 2) = \frac{P(X = 2 | Y = 3)}{P(X = 2)}$

From the attached image, we have:

$P(X = 2 | Y = 3) = 0.10$ i.e. column x = 2 and row y = 3

$P(X = 2) = 0.05 + 0.10 + 0.20$ --- i.e. we add all columns of x = 2

This gives:

$P(X = 2) = 0.35$

So, we have:

$P(Y = 3 | X = 2) = \frac{P(X = 2 | Y = 3)}{P(X = 2)}$

$P(Y = 3 | X = 2) = \frac{0.10}{0.35}$

$P(Y = 3 | X = 2) = 0.2857$