Question

A certain circuit board consists of two resistors, green and red. The circuit board manufacturer has two huge bins filled with the resistors, one for each color. Based on several years of data, it is known that 90% of the red resistors are functional, and 75% of the green resistors are functional. When creating a circuit board, the technician selects one red and one green resistor at random.(a) The circuit board as a whole is only functional if both resistors are functional. What is the probability that the circuit board is functional? (b) What is the probability that exactly one of the resistors chosen is functional?

Answer 1

Answer:

(a) 0.675

(b) 0.3

Step-by-step explanation:

Answer 2

Answer:

(a) 0.675

(b) 0.3

Step-by-step explanation:

$P(R) = 90\% = 0.9$ (Probability of functional red)

$P(\sim R) = 90\% = 1 - 0.9 = 0.1$ (Probability of non-functional red)

$P(G) = 75\% = 0.75$ (Probability of functional green)

$P(\sim G) = 1 - P(G) = 1 - 0.75 = 0.25$ (Probability of non-functional green)

(a) The probability the board is functional is the probability that the red and the green resistors are functional.

$P(\text{board is functional}) = P(R\cap G) = P(R) \times P(G) = 0.9\times0.75 = 0.675$

(b) The probability that exactly one of the resistors chosen is functional is the probability that either red is functional and green is not or green is functional and red is not.

$P(\text{exactly one is functional}) = P((R\cap \sim G) \cup (G\cap \sim R))$

$P(\text{exactly one is functional}) = P(R\cap \sim G) + P(G\cap \sim R) = (P(R) \times P(\sim G)) + (P(G) \times P(\sim R))$

$P(\text{exactly one is functional}) = (0.9 \times 0.25) + (0.75 \times 0.1) = 0.225 + 0.075 = 0.3$