Four percent of DVD players manufactured by an electronics company are defective. Suppose six DVD players are randomly selected from the production line. Each DVD player is determined to be defective or non-defective.

a. What is the probability that none of the DVD players are defective?

b. What is the probability that all six of the DVD players are defective?

c. What is the probability that at most one of the DVD players is defective?

d. What is the probability that at least two of the DVD players are defective?

Pr(all are good)=.96^6

Pr(six bad)=.04^6
Pr(five+good)=.04 *.96^5 + .96^6
Pr(two or more defiective)=Pr(all good or one defective)=.96^6+.96*.04^5

To solve these probability questions, we can use the concept of the binomial distribution. The binomial distribution is used when there are two possible outcomes (defective or non-defective) for each trial (selecting a DVD player), and the trials are independent and have the same probability of success (defective) for each trial.

The probability of success (defective) is given as 4% or 0.04, while the probability of failure (non-defective) is 1 minus the probability of success, which is 1 - 0.04 = 0.96.

Now let's answer each question:

a. What is the probability that none of the DVD players are defective?

To find this probability, we need to find the probability of getting a non-defective DVD player for all six trials. The probability of getting a non-defective DVD player for one trial is 0.96. Since the trials are independent, we multiply the probabilities together:

P(none defective) = 0.96 * 0.96 * 0.96 * 0.96 * 0.96 * 0.96 ≈ 0.817

Therefore, the probability that none of the DVD players are defective is approximately 0.817 or 81.7%.

b. What is the probability that all six of the DVD players are defective?

To find this probability, we need to find the probability of getting a defective DVD player for all six trials. The probability of getting a defective DVD player for one trial is 0.04. Since the trials are independent, we multiply the probabilities together:

P(all defective) = 0.04 * 0.04 * 0.04 * 0.04 * 0.04 * 0.04 ≈ 0.000004096

Therefore, the probability that all six of the DVD players are defective is approximately 0.000004096 or 0.0004%.

c. What is the probability that at most one of the DVD players is defective?

To find this probability, we need to find the probabilities of having exactly 0 defective players and exactly 1 defective player, and then sum them together.

P(at most one defective) = P(none defective) + P(one defective)

We already calculated P(none defective) to be approximately 0.817.

To find P(one defective), we need to consider all the possible ways of having exactly one defective player in six trials. There are C(6,1) = 6 ways to choose which DVD player is defective, and for each of these choices, the probability of having one defective player is:

P(one defective) = 6 * (0.04 * 0.96 * 0.96 * 0.96 * 0.96 * 0.96)

The final probability is the sum of P(none defective) and P(one defective).

P(at most one defective) = 0.817 + 6 * (0.04 * 0.96^5) ≈ 0.913

Therefore, the probability that at most one of the DVD players is defective is approximately 0.913 or 91.3%.

d. What is the probability that at least two of the DVD players are defective?

To find this probability, we need to find the probabilities of having exactly two defective players, exactly three defective players, and so on, up to exactly six defective players, and then sum them together.

P(at least two defective) = P(two defective) + P(three defective) + P(four defective) + P(five defective) + P(six defective)

To find each of these probabilities, we use the same logic as in part c, but summing the probabilities for each case:

P(two defective) = C(6,2) * (0.04^2 * 0.96^4)
P(three defective) = C(6,3) * (0.04^3 * 0.96^3)
P(four defective) = C(6,4) * (0.04^4 * 0.96^2)
P(five defective) = C(6,5) * (0.04^5 * 0.96)
P(six defective) = C(6,6) * (0.04^6)

The final probability is the sum of these individual probabilities.

P(at least two defective) = P(two defective) + P(three defective) + P(four defective) + P(five defective) + P(six defective)

After calculating the values, we find:

P(at least two defective) ≈ 0.9982

Therefore, the probability that at least two of the DVD players are defective is approximately 0.9982 or 99.82%.