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A town recently dismissed 5 employees in order to meet their new budget reductions. The town...

A town recently dismissed 5 employees in order to meet their new budget reductions. The town had 7 employees over 50 years of age and 16 under 50. If the dismissed employees were selected at random, what is the probability that no more than 1employee was over 50? Express your answer as a fraction or a decimal number rounded to four decimal places.

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This is a combination problem, so we work with different possible combinations to get the required probability.

we have to dismiss 5 employees

7 employees are over 50 years and 16 employees under 50

total number of employees = 7+16 = 23

Number of ways of selecting 0 employee over 50 = combination(7,0)*combination(16,5) = (7!/(7-0)!*0!)*(16!/(16-5)!*5!)

this gives us

Number of ways of selecting 0 employee over 50 = 4368 ways

Similarly, number of ways of dismissing 1 employee over 50 = combination(7,1)*combination(16,4) = (7!/(7-1)!*1!)*(16!/(16-4)!*4!)

this gives us

Number of ways of dismissing 1 employee over 50 = 12740 ways

adding both number of ways of dismissing 0employee over 50 years and 1 employee over 50 years = 4368+12740 = 17108

and total number of ways of dismissing 5 employee out of 23 employees= combination(23,5) = (231/(23-5)! * 5!) = 33649

So, required probability = (number of ways dismissing 0 and 1 employee over 50 years)/(total number of ways)

setting the calculated values, we get

Required probability = (17108/33649) = 0.5084

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