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In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
- a)5/21
- b)3/7
- c)4/7
- d)5/7
- e)16/21
Correct answer is option 'E'. Can you explain this answer?
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Verified Answer
In a room filled with 7 people, 4 people have exactly 1 friend in the ...
Begin by counting the number of relationships that exist among the 7 individuals whom we will call A, B, C, D, E, F, and G.
First consider the relationships of individual A: AB, AC, AD, AE, AF, AG = 6 total. Then consider the relationships of individual B without counting the relationship AB that was already counted before: BC, BD, BE, BF, BG = 5 total. Continuing this pattern, we can see that C will add an additional 4 relationships, D will add an additional 3 relationships, E will add an additional 2 relationships, and F will add 1 additional relationship. Thus, there are a total of 6 + 5 + 4 + 3 + 2 + 1 = 21 total relationships between the 7 individuals.
Alternatively, this can be computed formulaically as choosing a group of 2 from
We are told that 4 people have exactly 1 friend. This would account for 2 "friendship" relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 friends. This would account for another 3 "friendship" relationships (e.g. EF, EG, and FG). Thus, there are 5 total "friendship" relationships in the group.
The probability that any 2 individuals in the group are friends is 5/21. The probability that any 2 individuals in the group are NOT friends = 1 – 5/21 = 16/21.
First consider the relationships of individual A: AB, AC, AD, AE, AF, AG = 6 total. Then consider the relationships of individual B without counting the relationship AB that was already counted before: BC, BD, BE, BF, BG = 5 total. Continuing this pattern, we can see that C will add an additional 4 relationships, D will add an additional 3 relationships, E will add an additional 2 relationships, and F will add 1 additional relationship. Thus, there are a total of 6 + 5 + 4 + 3 + 2 + 1 = 21 total relationships between the 7 individuals.
Alternatively, this can be computed formulaically as choosing a group of 2 from
We are told that 4 people have exactly 1 friend. This would account for 2 "friendship" relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 friends. This would account for another 3 "friendship" relationships (e.g. EF, EG, and FG). Thus, there are 5 total "friendship" relationships in the group.
The probability that any 2 individuals in the group are friends is 5/21. The probability that any 2 individuals in the group are NOT friends = 1 – 5/21 = 16/21.
Most Upvoted Answer
In a room filled with 7 people, 4 people have exactly 1 friend in the ...
To solve this problem, we can use the concept of combinations and probability.
Step 1: Calculate the total number of ways to choose 2 individuals from a room of 7 people.
The total number of ways to choose 2 individuals from a group of 7 can be calculated using the combination formula: nCk = n! / (k!(n-k)!), where n is the total number of people and k is the number of people to be chosen.
In this case, n = 7 and k = 2, so the total number of ways to choose 2 individuals from the room is 7C2 = 7! / (2!(7-2)!) = (7*6) / (2*1) = 21.
Step 2: Calculate the number of ways to choose 2 individuals who are friends.
Let's consider the two cases separately:
Case 1: Choosing 2 individuals who have exactly 1 friend in the room.
There are 4 people who have exactly 1 friend in the room. To choose 2 individuals who are friends, we can choose one person from this group of 4, and then choose their friend from the remaining 3 people. This can be done in 4*3 = 12 ways.
Case 2: Choosing 2 individuals who have exactly 2 friends in the room.
There are 3 people who have exactly 2 friends in the room. To choose 2 individuals who are friends, we can choose one person from this group of 3, and then choose their friend from the remaining 2 people. This can be done in 3*2 = 6 ways.
Therefore, the total number of ways to choose 2 individuals who are friends is 12 + 6 = 18.
Step 3: Calculate the number of ways to choose 2 individuals who are NOT friends.
To calculate the number of ways to choose 2 individuals who are NOT friends, we subtract the number of ways to choose 2 individuals who are friends from the total number of ways to choose 2 individuals.
So, the number of ways to choose 2 individuals who are NOT friends is 21 - 18 = 3.
Step 4: Calculate the probability of choosing 2 individuals who are NOT friends.
The probability is calculated by dividing the number of favorable outcomes (2 individuals who are NOT friends) by the total number of possible outcomes (2 individuals chosen from the room).
So, the probability = number of favorable outcomes / total number of outcomes = 3 / 21 = 1 / 7.
Therefore, the probability of choosing two individuals who are NOT friends is 1/7, which is equivalent to option 'E' (16/21) in the given options.
Step 1: Calculate the total number of ways to choose 2 individuals from a room of 7 people.
The total number of ways to choose 2 individuals from a group of 7 can be calculated using the combination formula: nCk = n! / (k!(n-k)!), where n is the total number of people and k is the number of people to be chosen.
In this case, n = 7 and k = 2, so the total number of ways to choose 2 individuals from the room is 7C2 = 7! / (2!(7-2)!) = (7*6) / (2*1) = 21.
Step 2: Calculate the number of ways to choose 2 individuals who are friends.
Let's consider the two cases separately:
Case 1: Choosing 2 individuals who have exactly 1 friend in the room.
There are 4 people who have exactly 1 friend in the room. To choose 2 individuals who are friends, we can choose one person from this group of 4, and then choose their friend from the remaining 3 people. This can be done in 4*3 = 12 ways.
Case 2: Choosing 2 individuals who have exactly 2 friends in the room.
There are 3 people who have exactly 2 friends in the room. To choose 2 individuals who are friends, we can choose one person from this group of 3, and then choose their friend from the remaining 2 people. This can be done in 3*2 = 6 ways.
Therefore, the total number of ways to choose 2 individuals who are friends is 12 + 6 = 18.
Step 3: Calculate the number of ways to choose 2 individuals who are NOT friends.
To calculate the number of ways to choose 2 individuals who are NOT friends, we subtract the number of ways to choose 2 individuals who are friends from the total number of ways to choose 2 individuals.
So, the number of ways to choose 2 individuals who are NOT friends is 21 - 18 = 3.
Step 4: Calculate the probability of choosing 2 individuals who are NOT friends.
The probability is calculated by dividing the number of favorable outcomes (2 individuals who are NOT friends) by the total number of possible outcomes (2 individuals chosen from the room).
So, the probability = number of favorable outcomes / total number of outcomes = 3 / 21 = 1 / 7.
Therefore, the probability of choosing two individuals who are NOT friends is 1/7, which is equivalent to option 'E' (16/21) in the given options.
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In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOTfriends?a)5/21b)3/7c)4/7d)5/7e)16/21Correct answer is option 'E'. Can you explain this answer?
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