Question 1200582: The number of sick days an employee takes in one year is uniformly distributed on the values 0,
1, 2, …,10 and is independent from one year to the next. In a five-year period, what is the
probability that an employee takes more than 5 sick days in at most one of the years?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Let's mark (in blue) when an employee takes more than 5 sick days.
0,1,2,3,4,5,6,7,8,9,10
0,1,2,3,4,5,6,7,8,9,10
There are 10-6+1 = 5 highlighted items out of 11 total.
5/11 is the probability an employee takes more than 5 sick days for any given year.
We can break things down into two events- A = employee takes 5 or less sick days
- B = employee takes more than 5 sick days
which suggests we might have a binomial situation here.
A binomial situation must satisfy the following conditions:- Two outcomes.
- Each trial is independent any other.
- The probability p stays the same for each trial.
We can see that- There are two outcomes (either event A happens, or B does)
- Any given year is independent of any other.
- The probability of 5/11 stays the same for any given year.
Therefore, we have confirmed this is indeed a binomial process.
n = 5 year period = sample size
p = 5/11 = probability of event B occurring
x = number of years where event B occurs
The value of x is from the set {0,1,2,3,4,5}
x = 0 means event B occurs 0 times, so event A occurs all 5 times
x = 1 means event B occurs 1 time, so event A occurs 4 times
x = 2 means event B occurs 2 times, so event A occurs 3 times
etc
Let's compute the binomial probability when x = 0
(n C x)*(p^x)*(1-p)^(n-x)
(5 C x)*((5/11)^x)*(1-5/11)^(5-x)
(5 C x)*((5/11)^x)*(6/11)^(5-x)
(5 C 0)*((5/11)^0)*(6/11)^(5-0)
(1)*((5/11)^0)*(6/11)^(5-0)
0.04828284208108
This value is approximate.
There's about a 4.83% chance of event B occurring exactly 0 times.
Now repeat for x = 1
(n C x)*(p^x)*(1-p)^(n-x)
(5 C x)*((5/11)^x)*(1-5/11)^(5-x)
(5 C x)*((5/11)^x)*(6/11)^(5-x)
(5 C 1)*((5/11)^1)*(6/11)^(5-1)
(5)*((5/11)^1)*(6/11)^(5-1)
0.20117850867117
This value is approximate.
There's about a 20.12% chance of event B occurring exactly 1 time.
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The results we got were
0.04828284208108
0.20117850867117
They add to
0.04828284208108 + 0.20117850867117 = 0.24946135075224
This rounds to about 0.2495
Round this final value however needed.
There's about a 24.95% chance of event B happening at most one time (i.e. 1 or fewer).
There's about a 24.95% chance of an employee taking more than 5 sick days at most one time.
Here are a few binomial distribution calculators
https://www.omnicalculator.com/statistics/binomial-distribution
https://www.gigacalculator.com/calculators/binomial-probability-calculator.php
and here is a resource for TI84 calculators
https://www.statology.org/binomial-probabilities-ti-84-calculator/
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Answer: 0.2495 (approximate)
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