Two fair dice are thrown. The number of cases where the number appearing on the upper face of the first die is not less than that on the lower face of the second die is
21
Dice Throw Outcomes
The problem asks for the number of cases when two fair dice are thrown, such that the number on the upper face of the first die is not less than the number on the upper face of the second die.
Let $D_1$ be the number shown on the upper face of the first die, and $D_2$ be the number shown on the upper face of the second die.
Each die can show a number from 1 to 6. Since the dice are fair, each outcome $(D_1, D_2)$ is equally likely. The total number of possible outcomes when throwing two dice is $6 \times 6 = 36$.
We are looking for the number of pairs $(D_1, D_2)$ such that the condition $D_1 \ge D_2$ is satisfied. "Not less than" means "greater than or equal to".
We can list the possible outcomes that satisfy this condition:
- If the first die shows 1 ($D_1=1$), the second die must show a number less than or equal to 1. The only possibility is $D_2=1$. The outcome is (1, 1). (1 case)
- If the first die shows 2 ($D_1=2$), the second die must show a number less than or equal to 2. The possibilities are $D_2 \in \{1, 2\}$. The outcomes are (2, 1), (2, 2). (2 cases)
- If the first die shows 3 ($D_1=3$), the second die must show a number less than or equal to 3. The possibilities are $D_2 \in \{1, 2, 3\}$. The outcomes are (3, 1), (3, 2), (3, 3). (3 cases)
- If the first die shows 4 ($D_1=4$), the second die must show a number less than or equal to 4. The possibilities are $D_2 \in \{1, 2, 3, 4\}$. The outcomes are (4, 1), (4, 2), (4, 3), (4, 4). (4 cases)
- If the first die shows 5 ($D_1=5$), the second die must show a number less than or equal to 5. The possibilities are $D_2 \in \{1, 2, 3, 4, 5\}$. The outcomes are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5). (5 cases)
- If the first die shows 6 ($D_1=6$), the second die must show a number less than or equal to 6. The possibilities are $D_2 \in \{1, 2, 3, 4, 5, 6\}$. The outcomes are (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6). (6 cases)
Counting Favorable Cases
The total number of cases where the number on the first die is not less than the number on the second die is the sum of the cases for each value of the first die:
\( \text{Total cases} = 1 + 2 + 3 + 4 + 5 + 6 \)
Calculating the sum:
\( 1 + 2 + 3 + 4 + 5 + 6 = 21 \)
So, there are 21 cases where the number on the upper face of the first die is not less than that on the upper face of the second die.
Alternatively, we can visualize this using a table of all 36 outcomes:
| \(D_1\) \(\downarrow\), \(D_2\) \(\rightarrow\) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | (1,1) \(\checkmark\) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
| 2 | (2,1) \(\checkmark\) | (2,2) \(\checkmark\) | (2,3) | (2,4) | (2,5) | (2,6) |
| 3 | (3,1) \(\checkmark\) | (3,2) \(\checkmark\) | (3,3) \(\checkmark\) | (3,4) | (3,5) | (3,6) |
| 4 | (4,1) \(\checkmark\) | (4,2) \(\checkmark\) | (4,3) \(\checkmark\) | (4,4) \(\checkmark\) | (4,5) | (4,6) |
| 5 | (5,1) \(\checkmark\) | (5,2) \(\checkmark\) | (5,3) \(\checkmark\) | (5,4) \(\checkmark\) | (5,5) \(\checkmark\) | (5,6) |
| 6 | (6,1) \(\checkmark\) | (6,2) \(\checkmark\) | (6,3) \(\checkmark\) | (6,4) \(\checkmark\) | (6,5) \(\checkmark\) | (6,6) \(\checkmark\) |
The checkmarks (\(\checkmark\)) indicate the outcomes where \(D_1 \ge D_2\). Counting the checkmarks, we find there are 21 such cases.
The number of cases where the number appearing on the upper face of the first die is not less than that on the upper face of the second die is 21.
Important Questions from Numerical Estimation
The number of digits you have to type to write all the page numbers of a book starting from I (first page) is 2019. What is the number of pages in that book?
The number of Hens, Ducks, and Goats in farm P are 65, 91 and 169, respectively. The total number of Hens, Ducks and Goats in a nearby farm Q is 416. The ratio of hens : ducks : goats in farm Q is 5 : 14 : 13. All the hens, ducks and goats are sent from farm Q to farm P.
The new ratio of hens : ducks : goats in farm P is ____________.
A person divided an amount of Rs. 100,000 into two parts and invested in two different schemes. In one he got 10% profit and in the other he got 12%. If the profit percentages are interchanged with these investments he would have got Rs.120 less. Find the ratio between his investments in the two schemes.
S, M, E and F are working in shifts in a team to finish a project. M works with twice the efficiency of others but for half as many days as E worked. S and M have 6 hour shifts in a day, whereas E and F have 12 hours shifts. What is the ratio of contribution of M to contribution of E in the project?
M and N start from the same location. M travels 10 km East and then 10 km North – East. N travels 5 km South and then 4 km South – East. What is the shortest distance (in km) between M and N at the end of their travel?
