For 3(1), I got (6*3*2*2*1*1)/(6!) for 3(2), my reasoning is that holds the same
ID: 3327939 • Letter: F
Question
For 3(1), I got (6*3*2*2*1*1)/(6!)
for 3(2), my reasoning is that holds the same logic from 3(1), yet combine additional 3 boys. Therefore I got (9*3*2*2*1*1*C(n=6,r=2))/(9!)
Did I make a right approach? If not, could you please enlighten me? Thank you!
3. (1)There are 6 chairs around a table. A group of 3 boys and 3 girls take the seats randomly. What is the probability that no girl sits beside another girl? (2)There are 9 chairs around a table. A group of 6 boys and 3 girls take the seats randomly. What is the probability that no girl sits beside another girl?Explanation / Answer
Yes you are make right approch, I appriciate your work,
1)
Probability that no girl sits beside another girl = (6*3*2*2*1*1)/(6!) = 72 /720 = 0.1
2)
Probability that no girl sits beside another girl = (9*3*2*2*1*1*C(n=6,r=2))/(9!)
= 108 * 15 / 362880
= 0.0045
Probability that no girl sits beside another girl = 0.0045
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.