Mathematics, 28.02.2020 01:30 genyjoannerubiera

A company hires management trainees for entry level sales positions. Past experience indicates that only 10% will still be employed at the end of nine months. Assume the company recently hired six trainees.

a. What is the probability that three of the trainees will still be employed at the end of nine months?
b. What is the probability that at least two of the trainees will still be employed at the end of nine months?

Answers: 3

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Answer from: arthkk0877

The probability will be:(a) 0.0416(b) 0.1143

According to the question,

Sample size,

n = 6

Probability of event,

p = 0.1

(a)

The probability,

→ $P(X = 3) = C(6,3)\times 0.1^3\times (1-0.1)^3$

= 0.0416

(b)

→ $P(X \geq 2) = 1-P(X =0)- P(X =1)$

= 0.1143

Thus the above answer is appropriate.

Learn more about probability here:

link

Answer from: alethhaaa

a) 1.46% probability that three of the trainees will still be employed at the end of nine months.

b) 11.43% probability that at least two of the trainees will still be employed at the end of nine months

Step-by-step explanation:

For each trainee, there are only two possible outcomes. Either they will still be employed at the end of nine months, or they will not. The probability of a trainer being employed is independent from other trainees. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

10% will still be employed at the end of nine months.

This means that p = 0.1

Assume the company recently hired six trainees.

This means that n = 6

a. What is the probability that three of the trainees will still be employed at the end of nine months?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{6,3}.(0.1)^{3}.(0.9)^{3} = 0.0146$

1.46% probability that three of the trainees will still be employed at the end of nine months.

b. What is the probability that at least two of the trainees will still be employed at the end of nine months?

Either less than two will be employed, or at least two will. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

$P(X \geq 2) = 1 - P(X < 2)$

In which

P(X < 2) = P(X = 0) + P(X = 1)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.1)^{0}.(0.9)^{6} = 0.5314$

$P(X = 1) = C_{6,1}.(0.1)^{1}.(0.9)^{5} = 0.3543$

P(X < 2) = P(X = 0) + P(X = 1) = 0.5314 + 0.3543 = 0.8857

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8857 = 0.1143$

11.43% probability that at least two of the trainees will still be employed at the end of nine months

Answer from: Quest

see explanation

step-by-step explanation:

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

remove brackets and collect like terms

13 - $\frac{3}{2}$ r - 1 + r

= 12 - $\frac{1}{2}$ r

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

remove brackets and collect like terms

= 7r - $\frac{3}{2}$ - $\frac{2}{3}$ - 6r

[- $\frac{3}{2}$ - $\frac{2}{3}$ = - $\frac{9}{6}$ - $\frac{4}{6}$ = - $\frac{13}{6}$ ]

= r - $\frac{13}{6}$

Answer from: Quest

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake

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