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Juliette [100K]
3 years ago
9

To transfer into a particular technical department, a company requires an employee to pass a screening test. A maximum of 3 atte

mpts are allowed at 6-month intervals between trials. From past records it is found that 40% pass on the first trial; of those that fail the first trial and take the test a second time, 60% pass; and of those that fail on the second trial and take the test a third time, 20% pass. For an employee wishing to transfer:
(A) What is the probability of passing the test on the first or second try?
(B) What is the probability of failing on the first 2 trials and passing on the third?
(C) What is the probability of failing on all 3 attempts?

Mathematics
1 answer:
kiruha [24]3 years ago
3 0

Answer:

a) 0.760

b) 0.048

c) 0.192

Step-by-step explanation:

The step by step solution is attached as an image.

A) The probability of passing the test on the first or second try is 0.760.

That is he pass in the first trial or second trial.

(B) The probability of failing on the first 2 trials and passing on the third is 0.048.

That is the employee fail the first the trial and pass the third trial.

(C) The probability of failing on all 3 attempts is 0.192.

That is the employee fail all the three trial.

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b. (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

Step-by-step explanation:

a. Why do we need imaginary numbers?

We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution. For example, the equation of the form x² + 2x + 1 = 0 has the solution (x - 1)(x + 1) = 0 , x = 1 twice. The equation x² + 1 = 0 has the solution x² = -1 ⇒ x = √-1. Since we cannot find the square-root of a negative number, the identity i = √-1 was developed to be the solution to the problem of solving quadratic equations which have the square-root of a negative number.

b. Expand (a + ib)⁵

(a + ib)⁵ =  (a + ib)(a + ib)⁴ = (a + ib)(a + ib)²(a + ib)²

(a + ib)² = (a + ib)(a + ib) = a² + 2iab + (ib)² = a² + 2iab - b²

(a + ib)²(a + ib)² = (a² + 2iab - b²)(a² + 2iab - b²)

= a⁴ + 2ia³b - a²b² + 2ia³b + (2iab)² - 2iab³ - a²b² - 2iab³ + b⁴

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collecting like terms, we have

= a⁴ + 2ia³b + 2ia³b - a²b² - 4a²b² - a²b² - 2iab³  - 2iab³ + b⁴

= a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴

(a + ib)(a + ib)⁴ = (a + ib)(a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴)

= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b + 4i²a³b² - 6ia²b³ - 4i²ab⁴ + ib⁵

= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b - 4a³b² - 6ia²b³ + 4ab⁴ + ib⁵

collecting like terms, we have

= a⁵ + 4ia⁴b + ia⁴b - 6a³b² - 4a³b² - 4ia²b³ - 6ia²b³ + ab⁴ + 4ab⁴ + ib⁵

= a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

So, (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

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