Question

1. Out of six computer chips, two are defective. If two of the chips are chosen randomly for testing, compute the probability that both of them are defective. List all the outcomes in the sample space, if we were checking all six computer chips.
2. A quiz consists of 6 multiple-choice questions. Each question has 4 possible answers. A student is unprepared, and he has no choice but to guess answers completely at random. He passes the quiz if he gets at least 3 questions correctly. What is the probability that he will pass?

The following description applies to problems 3 and 4.
A computer program is tested by 5 independent tests. If there is an error, these tests will discover it with probabilities 0.1, 0.2, 0.3, 0.4, and 0.5 respectively. Suppose that the program contains an error. What is the probability that it will be found

3. by at least one test? (Hint: this event is the complement of the event where the error is not found)
4. by all five tests?

Answer 1

Since out of 6 chips, 2 are defective so the probability of getting one chip out of 2 defective is 2 /6. After that out of 5 remaining chips, 1 is defective so probability of getting both defective is

P(both defective) = (2/6) * (1/5) = 0.0667

Let D shows the defective chip and G shows the good chip so possible sample space is

S = { DD, DG, GD, GG}

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Since each questions has 4 choices so probability of getting correct answer is

p = 1/4= 0.25

Let X is a random variable shows the number of correct answers out of 6. Here X has binomial distribution with parameters n=6 and p=0.25.

The probability that he will pass is

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The probability that no test will found error is

(1-0.1)*(1-0.2)*(1-0.3)*(1-0.4)*(1-0.5) = 0.1512

The  probability that it will be found by at least one test is

1 - 0.1512 = 0.8488