The problem asks for the probability that when rolling two identical cubic dice, at least one die shows a face value greater than 3.
A standard cubic die has faces numbered 1, 2, 3, 4, 5, 6.
It's often easier to calculate the probability of the event *not* happening and subtract it from 1.
The event we want is "at least one face value > 3".
The complement event is "neither face value is greater than 3", which means "both face values are less than or equal to 3".
The probability of the complement event (both dice $\le 3$) is:
$ P(\text{both} \le 3) = \frac{\text{Number of outcomes where both} \le 3}{\text{Total possible outcomes}} = \frac{9}{36} = \frac{1}{4} $Now, we can find the probability of the original event (at least one face value > 3) using the complement rule:
$ P(\text{at least one} > 3) = 1 - P(\text{both} \le 3) $ $ P(\text{at least one} > 3) = 1 - \frac{1}{4} $ $ P(\text{at least one} > 3) = \frac{3}{4} $The probability that at least one of the face values is greater than 3 is $\frac{3}{4}$.
Three dice are thrown. What is the probability of getting a sum which is a perfect square?
Two distinct natural numbers from 1 to 9 are picked at random. What is the probability that their product has 1 in its unit place?
Two dice are thrown. What is the probability that difference of numbers on them is 2 or 3 ?
Suppose that there is a chance for a newly constructed building to collapse, whether the design is faulty or not. The chance that the design is faulty is 10%. The chance that the building collapses is 95% if the design is faulty, otherwise it is 45%. If it is seen that the building has collapsed, then what is the probability that it is due to faulty design?
What is the probability that all three boys sit together?