A computer program is tested by 5 independent tests. If there is an error, these tests will discover it with probabilities 0.1, 0.2, 0.3, 0.4, and 0.5, respectively. Suppose that the program contains an error. What is the probability that it will be found

Answer :

Answer:

0.8488

Explanation:

Let E =error found by test 1

Let F=error found by test 2

Let G=error found by test 3

Let H=error found by test 4

Let I= error found by test 5

Given P(E)=0.1, P(F)=0.2, P(G)=0.3, P (H)= 0.4, P (I)=0.5

therefore P(notE)=0.9, P(notF)=0.8, P(notG)=0.7, P(not H)=0.6, P (notI)=0.5

Tests are independent P(not E & not F &not G & not H & not I=P(notE)*P(notF)*P(notG)*P (notH)*P (not I) =0.9*0.8*0.7*0.6*0.5 =0.1512

P(found by at least one test)= 1- P(not found by any test)=1-P(not E& not F & not G & not H & not I ) = 1-0.1512 = 0.8488

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