Probability problem

2098 views 2 replies

1. a packet of 10 electronic components is known to include 3 defectives.if 4 components are selected from the packet at random,what is the expected value of the number of defective? ans is 1.20,how?

 

2. the probability that there is at least one error in an account statement prepared by 3 persons A,B & C are 0.2,0.3,0.1 respectively.if A,B,C prepare 60,70,90 such statements,then the expected number of correct statements ? ans is 178,how?

 

3. a bag contains 6 white and 4 red balls.if a person draws 2 balls & receives 10/- & 20/- for a white & red balls respectively,then the expected amount is? ans is 28/-,how?

Replies (2)

1)4*3/10 = 1.2

2)(60*0.8)+(70*0.7)+(90*0.9)=178

3) {(0.6+0.56)*20 + (0.6+0.44)*30 + (0.4+0.33)*40 + (0.4+0.67)*30} / 4

=(23.2 + 31.2 + 29.2 + 32.1) / 4

=28.9 

not sure about third problem. let others comment

Answer - 3

1. If both the white balls drawn = 6C2 / 10C2 = 15/45

2. If both the red balls drawn    = 4C2 / 10C2 = 6/45

3. If one white & one red drawn = (6C1*4C1) / 10C2 = 24/45

Expected amount = (15/45*20) + (6/45*40) + (24/45*30) = 28

Note :- here 20 = 10+10 i.e. 2 white balls

                      40 = 20+20 i.e. 2 red balls

                      30 = 10+20 i.e. 1 red & 1white

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