Probability Distribution Mean (Expectation), Variance :: Problems

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