bir su deposu 12 kova su ile doluyor tam dolu olan su deposunun içindeki 5 kovası kullanılıyor depoda kaç kova su kalır
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We seems to be being asked the probability of rolling at least one four in fifty tries, which should be pretty high. As usual with "at least" questions, we write
p = P(at least one 4) = 1 - P(no fours at all)
The probability of a four on one roll is of course 1/6, so the probability for "not a four" is 1-1/6=5/6. We want the probability of a conjunction of 50 independent events: no four on the first roll and no four on the second roll and...
For a conjunction of independent events we multiply their probabilities. That's 5/6 times itself 50 times. So our answer is
If we need a decimal approximation, it's