please help

[rouzism]

From a conveyor one after another 4 details arrive, the probability of product defect of each of which is equal to 0,1. The details are checked until the order of receipt, until 2 suitable details are recruited or until they are finished. Find the order of distribution, the mathematical expectation and the variance of a random variable X - the number of checked details.

 

[Deleted member 4993]

From a conveyor one after another 4 details arrive, the probability of product defect of each of which is equal to 0,1. The details are checked until the order of receipt, until 2 suitable details are recruited or until they are finished. Find the order of distribution, the mathematical expectation and the variance of a random variable X - the number of checked details.

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[rouzism]

I started with the values being 2, 3 and 4.
2 have been checked and they are suitable
3 have been checked and 2 are suitable
4 have been checked and 2 are suitable

P(X=2)=(1-0,1)*(1-0,1)=0,81
P(X=3)=2*0,9*0,9*0,1 =0,162
P(x=4)=3*0,9**0,9*0,1*0,1=0,0243
0,81+0,162+0,0243 = 0,9963 but to be right the sum has to be 1
I am wrong with something but i dont know with what
thats where i am stuck

 

[Dr.Peterson]

Have you included the case that 4 are checked, but fewer than 2 are suitable?