Question

Question 1: Basic Probability Theory A manufacturing facility is responsible for producing two components that are produced at states 1 (poor), 2 (below average), 3 (average), 4 (good), and 5 (excellent). For the final product (con sisting of both components) to be acceptable the sum of the two states needs to be at least equal to 6. At the mmen, the manufacturing processes are random, and hence each component is equally likely to be at any of the 5 states. Answer the following questions (a) What is the probability that a final product is acceptable? What is the probability that a final product is unacceptable? (b) What is the probability that a final product is acceptable given that the first component is produced at a state of 2? (c) Based on your answers on (a) and (b), are the two events (acceptable final product and first component state) independent? (d) What is the probability that the second component was produced at a state of 3 given that the final product is acceptable? Answer:

Answer 1

Answer:

Let X denote the state of component 1 and Y denote the state of component 2. Now the pmf of X and pmf of Y are given by :

$\small P(X=x) = \frac{1}{5} \; \; \; \; ; x=1,2,3,4,5 \newline P(Y=y) = \frac{1}{5} \; \; \; \; ; y=1,2,3,4,5$

(a)

We are asked to calculate the following probability :

P(final product is acceptable) = $\small P(X+Y \geq 6)$

$\small =\sum_{x=1}^5 P(X=x)P(X+Y \geq 6|X=x) \newline = \sum_{x=1}^5 \frac{1}{5} *P(x+Y \geq 6) \newline = \frac{1}{5} \sum_{x=1}^5 P(Y \geq 6-x) \newline = \frac{1}{5}(P(Y \geq 5) + P(Y \geq 4) + P(Y \geq 3) + P(Y \geq 2) + P(Y \geq 1)) \newline$

$\small \text{Now we observe that :} \newline P(Y \geq 5 ) = P(Y=5) =\frac{1}{5} \newline P(Y \geq 4 ) = P(Y=5) + P(Y=4) =\frac{2}{5} \newline P(Y \geq 3 ) = P(Y=5) + P(Y=4) +P(Y=3)=\frac{3}{5} \newline P(Y \geq 2 ) = P(Y=5) + P(Y=4) +P(Y=3) + P(Y=2)=\frac{4}{5} \newline P(Y \geq 1 ) = P(Y=5) + P(Y=4) +P(Y=3) + P(Y=2) + P(Y=1)=\frac{5}{5} \newline$

Thus,

P(final product is acceptable) = $\small \frac{1}{5}(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}+\frac{5}{5})$

$\small =\frac{1}{5}(\frac{15}{5}) \newline = \frac{3}{5}$

Now, P(final product is unacceptable) = 1 - P(final product is acceptable) = 2/5.

(b)

P(final product is acceptable | X = 2)

$\small = P(X+Y \geq 6 | X=2) \newline =P(2+Y \geq 6) \newline =P(Y \geq 4) \newline =\frac{2}{5}$

(c)

Clearly, $\small P(X+Y \geq 6 | X=2) \neq P(X+Y \geq 6)$

Thus, X+Y and X are not independent and thus we conclude that acceptable final product and first component state are not independent.

(d)

We need to find the following probability :

$\small P(Y=3 | X+Y \geq 6) =\frac{P(Y=3 \; \cap \; X+Y \geq 6)}{P(X+Y \geq 6)} \newline =\frac{P(Y=3)*P(X+Y \geq 6 | Y=3)}{P(X+Y \geq 6)} \newline = \frac{P(Y=3)*P(X+3 \geq 6 )}{P(X+Y \geq 6)} \newline = \frac{P(Y=3)*P(X \geq 3 )}{P(X+Y \geq 6)} \newline = \frac{1/5*3/5}{3/5} \newline =1/5$