Examples on probability, Mathematics

1. A machine comprises of three transformers A, B and C. Such machine may operate if at least 2 transformers are working. The probability of each transformer working is given as displayed below;

            P (A) = 0.6,      P (B) = 0.5,      P (C) = 0.7

A mechanical engineer went to inspect the working situations of those transformers. Determine the probabilities of having the given outcomes

i. Only one transformer operating

ii. Two transformers are operating

iii. All three transformers are operating

iv. None is operating

v.  At least two are operating

vi. At most two are operating

Solution

P(A) =0.6                     P(A) = 0.4                    P(B) = 0.5                    P(~B) = 0.5

P(C) = 0.7                    P(C¯) = 0.3

 i.  P(only one transformer is operating) is described by the given possibilities

                         1st                   2nd                  3rd

            P          (A                     B¯                      C¯ )        =  0.6 x 0.5 x 0.3 = 0.09

            P          (A                     B                      C¯ )         = 0.4 x 0.5 x 0.3 = 0.06

            P          (A                     B¯                      C)         = 0.4 x 0.5 x 0.7 = 0.14

∴ P(Only one transformer working)

            = 0.09 + 0.06 + 0.14 = 0.29

ii.  P(only two transformers are operating) is described by the given possibilities. 

                        1st                   2nd                  3rd

            P          (A                     B                      C¯ )        =  0.6 x 0.5 x 0.3 = 0.09

            P          (A                     B¯                     C)          = 0.6 x 0.5 x 0.7 = 0.21

            P          (A                     B                      C)         = 0.4 x 0.5 x 0.7 = 0.14

∴ P(Only two transformers are operating)

 = 0.09 + 0.21 + 0.14 = 0.44

iii.  P(all the three transformers are operating). 

 = P(A) x P(B) x P(C)

 = 0.6 x 0.5 x 0.7

= 0.21

iv.  P(none of the transformers is operating). 

= P(A) x P(B¯) x P(C¯)

= 0.4 x 0.5 x 0.3

= 0.06

v.  P(at least 2 working). 

 = P(exactly 2 working) + P(all three working)

  =  0.44 + 0.21

= 0.65

vi.  P(at most 2 working). 

     =  P(Zero working) + P(one working) + P(two working)

     = 0.06 +  0.29 + 0.44

= 0.79

Posted Date: 2/20/2013 4:31:15 AM | Location : United States







Related Discussions:- Examples on probability, Assignment Help, Ask Question on Examples on probability, Get Answer, Expert's Help, Examples on probability Discussions

Write discussion on Examples on probability
Your posts are moderated
Related Questions
Solve the fractional equation: Example: Solve the fractional equation 1/(x-2) +1/(x+3) =0 Solution: The LCD is (x - 2)(x + 3); therefore, multiply both sides of t

Proof of Constant Times a Function: (cf(x))′ = cf ′(x) It is very easy property to prove using the definition given you a recall, we can factor a constant out of a limit. No

Mike sells on the average 15 newspapers per week (Monday – Friday). Find the probability that 2.1 In a given week he will sell all the newspapers

Describe Three Ways to Write Negative Fractions? There are three different ways that a negative fraction can be written. They are all represent the same value. 1. The negative

Sharon needs to make 25 half-cup servings of soup. How many ounces of soup does she required? One cup is 8 ounces, so half a cup is 4 ounces. Multiply 25 by 4 ounces to find ou

Explain Measurement Conversions in details? The following tables show measurements of length, distance, and weight converted from one system to the other. Length and Distanc

Dividing Mixed Numbers Dividing mixed numbers is a 3-step process: 1. Convert the mixed numbers to improper fractions. 2. Divide the fractions 3. Convert the result ba

Determine the function f ( x ) .             f ′ ( x )= 4x 3 - 9 + 2 sin x + 7e x , f (0) = 15 Solution The first step is to integrate to fine out the most general pos


How many types of ogives?