mathewmithun wrote:I am getting stumped by this question-I wrote the different possibilities and I was getting 3/5. But clearly my approach was wrong and I am certain that my answer is also wrong. So pls help me solve this question.
Diana is going on a school trip along with her two brothers bruce and clark. 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 one of her brothers?
A: 1/27 B: 4/27 C: 5/27 D: 4/9 E: 5/9
OA is E
P(at least 1 brother is assigned to Diana's group) = 1 - P(neither brother is assigned to Diana's group).
It is given that Diana is assigned to one of the 3 groups.
What we need to determine is how Diana's brothers can be assigned RELATIVE to Diana's group.
Since the students are RANDOMLY assigned to the 3 groups, the probability of being assigned to Diana's group = 1/3.
P(neither brother is assigned to Diana's group):
P(the 1st brother is to NOT ASSIGNED to Diana's group) = 2/3. (Of the 3 groups, 2 do not include Diana.)
P(the 2nd brother is NOT ASSIGNED to Diana's group) = 2/3. (Of the 3 groups, 2 do not include Diana).
Since we want both events to happen, we multiply the fractions:
2/3 * 2/3 = 4/9.
P(at least 1 brother is assigned to Diana's group):
1 - 4/9 = 5/9.
The correct answer is
E.
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