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among employees of a certain company, 60% know c/c , 50% know r, 30% know both languages, and 20% know neither language.

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Explanation:

Let's define:

- C/C as event C

- R as event R

We are given:

P(C) = 0.6

P(R) = 0.5

P(C ∩ R) = 0.3

P(Neither C nor R) = 0.2

We can start by using the formula for the union of two events:

P(C∪ R) = P(C) + P(R) - P(C ∩R)

P(C∪ R) = 0.6 + 0.5 - 0.3 = 0.8

So, we know that 80% of the employees at the company knows at least one language (C or R).

Now, we can use this result with the information about those who don't know any language to solve for the remaining probabilities.

P(Neither C nor R) = P[(C ∪ R)^c]

= 1 - P(C ∪ R)

= 1 - 0.8

= 0.2

We can now subtract P(C ∩ R) from P(C) and P(R) to get the probability that someone only knows one language:

P(Only C) = P(C) - P(C ∩ R) = 0.6 - 0.3 = 0.3

P(Only R) = P(R) - P(C ∩ R) = 0.5 - 0.3 = 0.2

So, we now have the probabilities of employees who know only C, only R, both C and R, or neither C nor R:

P(Only C or Only R) = P(Only C) + P(Only R) = 0.3 + 0.2 = 0.5

P(Both C and R) = P(C ∩ R) = 0.3

P(Neither C nor R) = 0.2

Therefore, 50% of employees know only C or only R, 30% of employees know both C and R, and 20% of employees know neither C nor R.

User Shireesh Asthana
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