Question

A company produces a certain product, and each unit of this product may have 3 different types of defects. Let Di, D2,Ds represent the three different kinds of defects.
Suppose further that for each unit produced P(D) = .07 P(D) = .12 P(Ds) = .05 P(D, U Ds) = .14 P(Din D2nDs) = .01

(a) What is the probability that a unit does not have a type 1 defect?

(b) What is the probability that a unit has both a type 2 and 3 defect?

(c) What is the probability that a unit has both a type 2 and 3 defect, but not a type 1 defect?

(d) What is the probability that a unit has at most two defects?

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) 0.88

b) 0.02

c) 0.01

d) 0.99

Step-by-step explanation:

Step one: State the given parameters

            $P(D_{1} ) = 0.12$                                   $P(D_{2} ) = 0.07$

           $P(D_{3} ) = 0.05$                                    $P (D_{1} U D_{2} ) = 0.13$

          $P(D_{1}n D_{2}n D_{3}) = 0.01$                        $P(D_{1} U D_{3}) = 0.14$

Step 2 : Obtain the probability that a unit does not have a type 1 defect

         $P(\frac{}{D_{1} })$ =  $1 -P(D_{1} )$

                    = $1 - 0.12$

                    = 0.88  

Step 3 : Obtain the probability that a unit has both type 2 and 3 defect?

          The probability of the unit having both type 2 and type 3 defect is denoted as $P(D_{2} n D_{3} )$

   This is calculated as

                    $P(D_{2}n D_{3}) =P(D_{2} ) + P(D_{3}) - P(D_{2} U D_{3})\\\\ = 0.07 + 0,05 - 0.13$

                    =   0.02

Therefore P(D_{2} n D_{3} ) = 0.02

Step 4 : Obtain the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect

                  Let $P(\frac{}{D_{1}} n D_{2} n D_{3} )$ denote the  probability that the unit has both a type 2 and type 3 ,but not a type 1 defect.

This can be calculated as follows :

                      $P(\frac{}{D_{1}} n D_{2} n D_{3} ) = P(D_{2} n D_{3}) - P(D_{1} n D_{2}nD_{3})$

                                               =   0.02 - 0.01

                                               =  0.01

Step 4 : Obtain the probability that a unit has at most two defects

               P(at most 2 defects)  = 1 - P(all three defects)

                                                  = $1- P(D_{1} n D_{2}nD_{3})$

                                                  =  1 - 0.01

                                                  = 0.99