is N following a binomial law?

Question

A box contains three transistors A, B and C. A has been manufactured by a machine producing 3% of defective transistors, B by a machine producing 5% of defective transistors and C by a machine producing 7% of defective transistors.

 

Ten transistors are choosen at random and with reset. Let N be the number of defective transistors among these ten.

 

1. Compute P(N=0)

2. Is N following a binomial law? If yes, give its parameters

Answer 1

Step 1: Find the Probability of Selecting a Defective Transistor

The box contains three transistors, A, B, and C, each equally likely to be selected.

Their defect probabilities are:

A: 3% = 0.03

B: 5% = 0.05

C: 7% = 0.07

Since each transistor has a 1/3 probability of being selected, we use the law of total probability:

P(Defective) = (1/3)(0.03) + (1/3)(0.05) + (1/3)(0.07)

P(Defective) = (0.03 + 0.05 + 0.07)/3

P(Defective) = 0.05

Therefore, each randomly selected transistor has a 5% probability of being defective.

The probability of selecting a nondefective transistor is:

P(Not Defective) = 1 - 0.05 = 0.95

Step 2: Calculate P(N = 0)

N represents the number of defective transistors among 10 selections.

N = 0 means none of the 10 selected transistors is defective.

Since selections are made independently with replacement, the probability that all 10 are nondefective is:

P(N = 0) = (0.95)^10

P(N = 0) ≈ 0.5987

Therefore, there is approximately a 59.87% probability of obtaining zero defective transistors.

Step 3: Does N Follow a Binomial Distribution?

Yes, under the intended interpretation that each selection is an independent draw from the same mixture of transistor production.

A binomial distribution requires four conditions:

1. A fixed number of trials.
2. Two possible outcomes for each trial: defective or nondefective.
3. Independent trials.
4. The same probability of success on every trial.

In this problem:

Number of trials: n = 10

Probability of a defective transistor: p = 0.05

Probability of a nondefective transistor: 1 - p = 0.95

Because all four conditions are satisfied:

N ~ Binomial(10, 0.05)

The general binomial probability formula is:

P(N = k) = C(10, k)(0.05)^k(0.95)^(10-k)

Here, k represents the number of defective transistors.

Final Answers

1. P(N = 0) = 0.95^10 ≈ 0.5987 = 59.87%
2. Yes, N follows a binomial distribution with parameters n = 10 and p = 0.05.

Important Clarification: This solution assumes the machine defect rates apply independently to each new selection. If A, B, and C are literally three fixed physical transistors whose defect statuses never change, repeatedly selecting them with replacement would not generally produce the binomial distribution above. The usual intended interpretation of this exercise is independent selections from the three production sources.