An ordinary die is tossed 4 times. What is the probability of getting a number greater than 4 at least once?

Hello!

Denote the event in question as A. Consider the complementary event not A: die is tossed 4 times and each result is 4 or less. The probability of A is 1-probability of not A: P(A)=1-P(not A). The probability of (not A) is simpler to compute.

The event not A consists of 4 events: that the first result is <=4, the second <=4, the third and the fourth. These events are independent: the next result has no relation with the previous one(s). Therefore the probability P(not A) is the product of four corresponding probabilities.

These probabilities are all the same: the probability that the result of one tossing is <=4. There are 6 possibilities, 1, 2, 3, 4, 5, 6 and 4 of them are "suitable": 1, 2, 3, 4. So this probability is by definition 4/6 = 2/3.

So P(not A) = 2/3*2/3*2/3*2/3 = 16/81 and P(A) = 1 - 16/81 = 65/81 `approx 0.80.`

The answer: the probability is 65/81.