Probability

watchkimberly

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Feb 3, 2021
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Can someone correct me if I am wrong?
At the end of each week, a generous teacher gives some candy to a few randomly selected students. There are 30 students in this teacher's class- 16 girls and 14 boys.
Question: To avoid appearing to favor one gender, the teacher will randomly choose 3 of the 16 girls and 3 of the 14 boys. How many possible groups of 6 can she choose?
My answer was 593,775 after doing 30P6/6!
My question is the next part, I'm not sure how to go about this:
To avoid appearing to favor one gender, the teach will randomly choose 3 of the 16 girls and 3 of the 14 boys. How many possible groups of 6 can she choose?
This is what I did: (16C3/3!)*(14C3/3!) = 203840, 203840/30*29*28*27*26*25/6! = 0.39010797 = 39.01
 
1st question you posted: To avoid appearing to favor one gender, the teacher will randomly choose 3 of the 16 girls and 3 of the 14 boys. How many possible groups of 6 can she choose?

2nd question you posted: To avoid appearing to favor one gender, the teach will randomly choose 3 of the 16 girls and 3 of the 14 boys. How many possible groups of 6 can she choose?


They are the same question. Please update your post. Thanks.


You wrote: 0.39010797 = 39.01. Do you really believe that?
 
At the end of each week, a generous teacher gives some candy to a few randomly selected students. There are 30 students in this teacher's class- 16 girls and 14 boys.
Question: If the teacher wanted to give candy to 6 of the students, how many possible groups of 6 students can she choose? My answer, 30C6/6! = 593,775
Question 2: To avoid appearing to favor one gender, the teach will randomly choose 3 of the 16 girls and 3 of the 14 boys. How many possible groups of 6 can she choose? My answer, (16C3/3!)*(14C3/3!) = 203840, 203840/30*29*28*27*26*25/6! = 0.39010797 = 39.01
Is this correct?
 
Can you please list the 39.01 cases.

Assume the 30 students are b1, b2, ..., b14, g1, g2, ..., g16.

After you list the 39.01 cases we will talk further.
 
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