Probability Distribution Mean (Expectation), Variance :: Problems

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The probability that there is at least one error is an accounts statement prepared by A is 0.2 and for B and C they are 0.25 and 0.4 respectively A, B, and C prepared 10, 16 and 20 statements respectively. Find the expected number of correct statements in all

 Expected no. of correct statements in all prepared by "A", "B" and "C" together : 32

 Solution

Let "x", "y" and "z" represent the number of correct statements prepared by "A", "B" and "C" respectively.

For "A"

In the experiment of testing whether a statement prepared by "A" is correct or not, there are two possible events,

• The event of the statement having at least one error
• The event of the statement having no errors

The probability that there is at least one error is an accounts statement prepared by A is 0.2
⇒ Probability that a statement prepared by "A"

• has atleast one error = 0.2   ⇒   P(Error) = 0.2
• has no error = 0.8   ⇒   P(No Error) = 0.8

Considering the two events of "Error" or "No Error" to be the only possibilities, they are exhaustive events
⇒ P("Error" ∪ "No Error") = 1       → (1)

Since there can either be atleast one error or there cannot be any error, the two events of "Error" or "No Error" are mutually exclusive
⇒ P("Error" ∩ "No Error") = 0
(Or) P("Error" ∪ "No Error") = P("Error") + P("No Error")       → (2)

From (1) and (2) we can write
P("Error" ∪ "No Error") = P("Error") + P("No Error") = 1
⇒ P("Error") + P("No Error") = 1
⇒ 0.2 + P("No Error") = 1
⇒ P("No Error") = 1 − 0.2
⇒ P("No Error") = 0.8

No. of repetitions of the experiment = No. of statements made by "A" tested
⇒ N = 10

Expected no. of correct statements prepared by "A"

= Expected frequency of occurance of the event "No Error" in "N" repetitions of
the experiment
= N × P(No Error)
= 10 × 0.8
= 8

⇒ E(x) = 8

Similarly

For "B"

Probability that a statement prepared by "B"

• has atleast one error = 0.25   ⇒   P(Error) = 0.25
• has no error = 0.75   ⇒   P(No Error) = 0.75

No. of repetitions of the experiment = No. of statements made by "B" tested
⇒ N = 16

Expected no. of correct statements prepared by "B"

= Expected frequency of occurance of the event "No Error" in "N" repetitions of
the experiment
= N × P(No Error)
= 16 × 0.75
= 12

⇒ E(y) = 12

Similarly

For "C"

Probability that a statement prepared by "C"

• has atleast one error = 0.4   ⇒   P(Error) = 0.4
• has no error = 0.6   ⇒   P(No Error) = 0.6

No. of repetitions of the experiment = No. of statements made by "C" tested
⇒ N = 20

Expected no. of correct statements prepared by "C"

= Expected frequency of occurance of the event "No Error" in "N" repetitions of
the experiment
= N × P(No Error)
= 20 × 0.6
= 12

⇒ E(z) = 12

Expected no. of correct statements in all prepared by "A", "B" and "C" together
 ⇒ E (x + y + z) = E (x) + E (y) + E (z) = 8 + 12 + 12 = 32

 Credit : Vijayalakshmi Desu