probability of technical success

Probability of passing a multiple question exam type 30 with five choices per question. The rate cut is 60%?
I know that one way to do it is: the sum of k = 18 to 30 of [C (30, k) * (0.2 ^ k) * (0.8 ^ (nk))] But I do not see a problem with the following argument. is a probability of 0.2 you can guess a correct question. need 60% of 30 is 18 right questions. The other questions are not important to have a chance to get 1 success of others. So just to mix the probability of guessing 18 times on a set of 30. This makes the equation: C (30.18) * (2 ^ 18) Is this wrong? If so please let me know some details. I'm decent at math, so do not worry about technique. Ps I'm more interested in the logic of the quadratic equation
It seems that something is missing. What if those 18 you? Then you do not pass the test. But What if one of the 12 questions that you are not taking into account that correct? Then go. So think you can not declare that the probable success for the other 12 questions is one. It seems that there are problems with his logic. Sorry, been a long time since I did Algebra 2, so I could be wrong. Good luck!
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