In a factory working a 5-day week, the probability of more than 10% of the employees being absent on any one day is 1/5. Calculate the probability that, in a given 5-day week, there are more than 10% absent  i) on exactly 2 days ii) on at least 2 days iii) on 2 successive days but on no other day

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You need to use binomial random variable which models the probability of getting exactly `k` successes in `n` Bernoulli trials. Probability density function of binomial random variable `X` is

`f(k;n,p)=P(X=k)=((n),(k))p^k(1-p)^(n-k)`                       (1)

where `p` i probability of success.

In your case success is when more than...

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You need to use binomial random variable which models the probability of getting exactly `k` successes in `n` Bernoulli trials. Probability density function of binomial random variable `X` is

`f(k;n,p)=P(X=k)=((n),(k))p^k(1-p)^(n-k)`                       (1)

where `p` i probability of success.

In your case success is when more than 10% of employees is absent hence `p=1/5`, `n=5` (5-day week).

i)

Here we only need to calculate probability density function for `k=2`.

`P(X=2)=((5),(2))(1/5)^2(4/5)^3=10cdot 1/25cdot64/125=640/3125=128/625`  <-- Solution

ii)

Here we will use probability of opposite event `P(A)=1-P(A^c)`. Opposite event of at least 2 days is less than 2 days i.e. more than 10% of employees is absent on 1 day or on 0 days. Hence

`P(X geq2)=1-P(X<2)=1-P(X=0)-P(X=1)=`

`1-((5),(0))(1/5)^0(4/5)^5-((5),(1))(1/5)(4/5)^4=1-1024/3125-256/625=821/3125`

`P(X geq2)=821/3125`  <--Solution

iii)

We will calculate this in a similar way as in i). The only difference is that 2 days must be successive. For this you must know what is the meaning of binomial coefficient in (1). It tells us in how many ways can we choose those two days (e.g. Monday and Tuesday or Monday and Wednesday or Tuesday and Friday etc.). In this case when we choose one day we automatically choose the next day as well (e.g. if we choose Monday we automatically chose Tuesday as well). So we calculate  probability the same as in i) but in this case our binomial coefficient is `((5),(1))`.

`P(X={2 "successive"})=((5),(1))(1/5)^2(4/5)^3=5cdot1/25cdot64/125=64/625`

`P(X={2 "successive"})=64/625`  <--Solution

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