The firm is requested to send 3 employees who have positive indications of asbestos on to a medical center for further testing. Suppose 40% of the employees have positive indications of asbestos in their lungs. a) Find the probability that exactly 10 employees will be tested in order to find 3 positives

0.0645 = 6.45% probability that exactly 10 employees will be tested in order to find 3 positives

Step-by-step explanation:

Binomial probability distribution

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

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

And p is the probability of X happening.

40% of the employees have positive indications of asbestos in their lungs

This means that

a) Find the probability that exactly 10 employees will be tested in order to find 3 positives

2 within the first 9( when ), and the 10th, with 0.4 probability. So

0.4*0.1612 = 0.0645

0.0645 = 6.45% probability that exactly 10 employees will be tested in order to find 3 positives

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)