Calculating Probability of Dice Rolls
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.
- Total possible outcomes for one die: 6
- Total possible outcomes when rolling two dice: $6 \times 6 = 36$.
Using the Complement Event Approach
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".
- Face values less than or equal to 3 are: {1, 2, 3}. There are 3 such values.
- Number of outcomes where the first die is $\le 3$: 3
- Number of outcomes where the second die is $\le 3$: 3
- Number of outcomes where *both* dice are $\le 3$: $3 \times 3 = 9$.
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} $Final Answer
The probability that at least one of the face values is greater than 3 is $\frac{3}{4}$.
Important Questions from Probability
Two dice are thrown simultaneously. What is the probability of getting the same number on both the dice?
The probability of being 53 Sundays in year 2020 is-
Three dice are thrown randomly. The probability of coming 3 in at least one die is
The probability of having 53 Tuesdays in an ordinary year is:
When two dice are tossed simultaneously, the probability that the sum of the numbers appearing on both the dice is 8 will be
