Is there anything wrong with the second part of this question if not how to solve

'In a shop there is a probability of 0.01 for faulty television to occur in stock.100 televisions are examined.By using binomial distribution find the expectation for the number of faulty television and the probability that more than one faulty television occur.
My answer for the first part is E(X)=np=100∙0.01=1. I'm now sure for the second part: I think more than one are faulty=X>1, and its opposite is at most one is good=X≤1. Hence P(X>1)=1-P(X≤1)=1-0∙C(0 100)0.99^0∙0.01^100-1∙C(1 100)0.99^1∙0.01^99=1 which means the probability for more than one faulty television to occur is 100%?!

for second part, you have garbled the formula

P[0] = .99^100 = .36603
P[1] = 100c1 *.01*.99^99 = .36973

P[>1] = 1 - ( .36603 + .36973) = .2642 <-----

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