1.An experiment consists of rolling two dice and the sum of faces is observed. How many observations are there (i.e. the possible outcomes)?
2. A bowl contains 10 red balls and 8 blue balls. If five balls are to be drawn, determine the number of ways that you can draw five blue balls?
3. Mike belongs to a group consisting of 10 members. One day, they decided to pose for a group pictorial with all of them standing in line. If you’re the photographer and Mark does not want to stand next to two particular members, in how many ways can you arrange them?
help me to answer this please with solution
(1,1), (1,2) (1,3), (1,4), (1,5),
(1,6). So, there 6 possibilities per side.
6poss./side * 6sides = 36 Possibilities.
1. To determine the number of observations or possible outcomes, we need to find all the possible sums that can be obtained from rolling two dice. Each die has 6 faces, numbered from 1 to 6. So, when we roll two dice, the sum of the faces can range from 2 (when both dice show a 1) to 12 (when both dice show a 6).
To calculate the number of observations, we count the number of possible sums. There are 11 possible sums: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.
So, there are 11 possible observations or outcomes when rolling two dice.
2. In this problem, we have a combination problem where we need to determine the number of ways to draw a certain number of blue balls from a total number of blue balls.
Given that there are 10 red balls and 8 blue balls in the bowl, we need to draw 5 balls. If we want to draw 5 blue balls, we need to choose all 5 balls from the blue ball pool.
To calculate the number of ways to draw 5 blue balls, we use the combination formula: nCr = n! / (r!(n-r)!), where n is the total number of balls and r is the number of balls to be drawn.
In this case, n = 8 (total number of blue balls) and r = 5 (balls to be drawn).
So, the number of ways to draw 5 blue balls is calculated as follows:
8C5 = 8! / (5!(8-5)!) = 8! / (5!3!) = (8 x 7 x 6 x 5 x 4) / (5 x 4 x 3 x 2 x 1) = 56
Therefore, there are 56 ways to draw 5 blue balls from the bowl.
3. In this problem, we need to arrange a group of 10 members in a line for a group pictorial. However, one member, Mark, does not want to stand next to two particular members.
To calculate the number of ways to arrange them, we can consider it as a permutation problem. We need to arrange 10 members, with Mark being one of them, in a line while ensuring the two particular members are not next to Mark.
We can approach this problem using the principle of inclusion-exclusion. Let's assume the two particular members are B1 and B2.
Step 1: Find the total number of ways to arrange all 10 members without any restrictions. This is simply the number of permutations of 10 objects, which is 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.
Step 2: Find the number of ways when B1 is next to Mark. In this case, we treat Mark and B1 as a single entity, so we have 9 objects to arrange (Mark&B1, B2, and the other 7 members). The number of ways to arrange these 9 objects is 9!.
Step 3: Find the number of ways when B2 is next to Mark. Similar to step 2, we treat Mark and B2 as a single entity, resulting in 9 objects (Mark&B2, B1, and the other 7 members) to arrange. The number of ways to arrange these 9 objects is again 9!.
Step 4: Find the number of ways when both B1 and B2 are next to Mark. In this case, we treat Mark, B1, and B2 as a single entity, so we have 8 objects to arrange (Mark&B1&B2 and the other 7 members). The number of ways to arrange these 8 objects is 8!.
To get the final answer, we use the principle of inclusion-exclusion. We subtract the sum of the cases from steps 2 to 4 from the total number of arrangements obtained in step 1.
Final answer = Total arrangements - (Arrangements with B1 next to Mark + Arrangements with B2 next to Mark - Arrangements with both B1 and B2 next to Mark)
Final answer = 10! - (9! + 9! - 8!)
These calculations will give you the number of ways you can arrange the group members while ensuring Mark does not stand next to the two particular members.