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Chapter 23: Problem 3

A machine manufactures 350 micro-chips per hour. The probability that a chip is faulty is \(0.012\). Calculate the probability that in a particular hour there are (a) one, (b) three, (c) more than three faulty chips manufactured.

Short Answer

Expert verified
Answer: The probabilities are as follows: a) Probability of one faulty chip manufactured: 0.0834 b) Probability of three faulty chips manufactured: 0.2025 c) Probability of more than three faulty chips manufactured: 0.4887

Step by step solution

01

Calculate the combinations

First, we need to calculate the combinations of selecting 1 success (faulty chip) out of 350 trials. $$ \binom{350}{1} = \frac{350!}{1!(350-1)!} = 350 $$
02

Calculate P(X = 1)

Next, we calculate the probability of one faulty chip manufactured using the Binomial Distribution formula. $$ P(X = 1) = \binom{350}{1}(0.012)^1 (1 - 0.012)^{349} = 350(0.012)(0.988)^{349} \approx 0.0834 $$ #b) Probability of three faulty chips manufactured#
03

Calculate the combinations

First, we need to calculate the combinations of selecting 3 successes (faulty chips) out of 350 trials. $$ \binom{350}{3} = \frac{350!}{3!(350-3)!} = 2044700 $$
04

Calculate P(X=3)

Next, we calculate the probability of three faulty chips manufactured using the Binomial Distribution formula. $$ P(X = 3) = \binom{350}{3}(0.012)^3 (1 - 0.012)^{347} = 2044700(0.012)^3(0.988)^{347} \approx 0.2025 $$ #c) Probability of more than three faulty chips manufactured#
05

Calculate P(X \(\leq\) 3)

To calculate the probability of more than three faulty chips manufactured, we need to find the complementary probability of at most three faulty chips, i.e. P(X \(\leq\) 3). Since we already calculated P(X = 1) and P(X = 3), we only need to calculate P(X = 2).
06

Calculate the combinations

First, we need to calculate the combinations of selecting 2 success (faulty chips) out of 350 trials. $$ \binom{350}{2} = \frac{350!}{2!(350-2)!} = 61225 $$
07

Calculate P(X=2)

Next, we calculate the probability of two faulty chips manufactured using the Binomial Distribution formula. $$ P(X=2) = \binom{350}{2}(0.012)^2(1-0.012)^{348} = 61225(0.012)^2(0.988)^{348} \approx 0.2254 $$
08

Calculate P(X > 3)

Now we can calculate the complementary probability of having more than three faulty chips manufactured. $$ P(X > 3) = 1 - (P(X = 1) + P(X = 2) + P(X = 3)) \approx 1 - (0.0834 + 0.2254 + 0.2025) \approx 0.4887 $$ In conclusion, the probabilities for each situation are: a) Probability of one faulty chip manufactured: 0.0834 b) Probability of three faulty chips manufactured: 0.2025 c) Probability of more than three faulty chips manufactured: 0.4887

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Most popular questions from this chapter

The probability that a motor will malfunction within 5 years of manufacture is \(0.03\). Out of eight motors calculate the probability that within 5 years of manufacture (a) all eight will malfunction (b) six will malfunction (c) none will malfunction.

Out of 6000 components, 39 fail within 12 months of manufacture. (a) Calculate the probability that a component picked at random fails within 12 months of manufacture. (b) A batch contains 2000 components. How many of these would you expect to fail within 12 months?

The probability that a component is acceptable is \(0.91\). Ten components are picked at random. Calculate the probability that (a) eight are acceptable (b) more than eight are acceptable (c) three are not acceptable.

The probability that a component fails within a month is \(0.009\). If 800 components are examined calculate the probability that the number failing within a month is (a) nine, (b) five, (c) less than three, (d) four or more.

The probability of passing a module on the first attempt is \(0.9\). A student takes six modules. Calculate the probability that the student (a) passes five modules (b) passes all modules (c) is required to take two or more resits.
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