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Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?
- a)1/27
- b)4/27
- c)5/27
- d)4/9
- e)5/9
Correct answer is option 'E'. Can you explain this answer?
Verified Answer
Diana is going on a school trip along with her two brothers, Bruce and...
Diana and her two brothers can be assigned to one of the 3 groups, so each has 3 choices, so total # of different assignments of Diana and her 2 brothers to 3 groups is 3∗3∗3 = 3 = 27.
Now, "Diana leaves at the same time as AT LEAST one her brothers" means that Diana is in the same group as at least one her brothers.
Let's find the opposite probability of such event and subtract it from 1. Opposite probability would be the probability that Diana is not in the group with any of her brothers.
In how many ways we can assign Diana and her two brothers to 3 groups so that Diana is not in the group with any of her brothers? If Diana is in the first group, then each of her two brothers will have 2 choices (either the second group or the third) and thus can be assigned in 2 ∗ 2 = 22 = 4 ways to other two groups. As there are 3 groups, then total # of ways to assign Diana and her 2 brothers to these groups so that Diana is not in the group with any her bothers is 3 ∗ 4 = 12 (For each of Diana's choices her brother can be assigned in 4 ways, as Diana has 3 choices: first, second or the third group, then total 3 ∗ 4 = 12). So probability of this event is 12/27.
Probability that Diana is in the same group as at least one her brothers would be
Answer: E.
Now, "Diana leaves at the same time as AT LEAST one her brothers" means that Diana is in the same group as at least one her brothers.
Let's find the opposite probability of such event and subtract it from 1. Opposite probability would be the probability that Diana is not in the group with any of her brothers.
In how many ways we can assign Diana and her two brothers to 3 groups so that Diana is not in the group with any of her brothers? If Diana is in the first group, then each of her two brothers will have 2 choices (either the second group or the third) and thus can be assigned in 2 ∗ 2 = 22 = 4 ways to other two groups. As there are 3 groups, then total # of ways to assign Diana and her 2 brothers to these groups so that Diana is not in the group with any her bothers is 3 ∗ 4 = 12 (For each of Diana's choices her brother can be assigned in 4 ways, as Diana has 3 choices: first, second or the third group, then total 3 ∗ 4 = 12). So probability of this event is 12/27.
Probability that Diana is in the same group as at least one her brothers would be
Answer: E.
Most Upvoted Answer
Diana is going on a school trip along with her two brothers, Bruce and...
To determine the probability that Diana leaves at the same time as at least one of her brothers, we can follow these steps:
Understanding the Scenario
- There are **3 students**: Diana (D), Bruce (B), and Clerk (C).
- Each student can be assigned to **3 different groups** (or times).
- The groups are **independent**, meaning assignments for each student do not affect others.
Total Possible Outcomes
- Each student has **3 choices** for the group they can join.
- The total number of ways to assign groups is:
\[
3 \times 3 \times 3 = 27
\]
Favorable Outcomes
To find the probability that Diana is in the same group as at least one brother:
1. **Calculate the Complement**:
- First, find the probability that Diana is **not** with either brother.
- If Diana is in a group, Bruce and Clerk each have **2 choices** (any group but Diana's).
- The number of outcomes where Diana is in a different group than both brothers is:
\[
3 \text{ (Diana's groups)} \times 2 \times 2 = 12
\]
2. **Calculate Favorable Outcomes**:
- The number of favorable outcomes (Diana with at least one brother) is:
\[
27 \text{ (total outcomes)} - 12 \text{ (Diana alone)} = 15
\]
Calculating Probability
- The probability that Diana is with at least one of her brothers is:
\[
P(\text{Diana with at least one brother}) = \frac{15}{27} = \frac{5}{9}
\]
Thus, the correct answer is option **E: 5/9**.
Understanding the Scenario
- There are **3 students**: Diana (D), Bruce (B), and Clerk (C).
- Each student can be assigned to **3 different groups** (or times).
- The groups are **independent**, meaning assignments for each student do not affect others.
Total Possible Outcomes
- Each student has **3 choices** for the group they can join.
- The total number of ways to assign groups is:
\[
3 \times 3 \times 3 = 27
\]
Favorable Outcomes
To find the probability that Diana is in the same group as at least one brother:
1. **Calculate the Complement**:
- First, find the probability that Diana is **not** with either brother.
- If Diana is in a group, Bruce and Clerk each have **2 choices** (any group but Diana's).
- The number of outcomes where Diana is in a different group than both brothers is:
\[
3 \text{ (Diana's groups)} \times 2 \times 2 = 12
\]
2. **Calculate Favorable Outcomes**:
- The number of favorable outcomes (Diana with at least one brother) is:
\[
27 \text{ (total outcomes)} - 12 \text{ (Diana alone)} = 15
\]
Calculating Probability
- The probability that Diana is with at least one of her brothers is:
\[
P(\text{Diana with at least one brother}) = \frac{15}{27} = \frac{5}{9}
\]
Thus, the correct answer is option **E: 5/9**.
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Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?a)1/27b)4/27c)5/27d)4/9e)5/9Correct answer is option 'E'. Can you explain this answer? for GMAT 2025 is part of GMAT preparation. The Question and answers have been prepared according to the GMAT exam syllabus. Information about Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?a)1/27b)4/27c)5/27d)4/9e)5/9Correct answer is option 'E'. Can you explain this answer? covers all topics & solutions for GMAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?a)1/27b)4/27c)5/27d)4/9e)5/9Correct answer is option 'E'. Can you explain this answer?.
Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?a)1/27b)4/27c)5/27d)4/9e)5/9Correct answer is option 'E'. Can you explain this answer? for GMAT 2025 is part of GMAT preparation. The Question and answers have been prepared according to the GMAT exam syllabus. Information about Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?a)1/27b)4/27c)5/27d)4/9e)5/9Correct answer is option 'E'. Can you explain this answer? covers all topics & solutions for GMAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?a)1/27b)4/27c)5/27d)4/9e)5/9Correct answer is option 'E'. Can you explain this answer?.
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