Of 27 students in a class, 10 have blue eyes , 14 have brown eyes , and

Question

3 have green eyes. If 3 students are chosen at random
a.what are the odds of all three have blue eyes?
b.what are the odds of NO one has brown eyes ?

Answer 1

first find the probability of all 3 blue eyes
= C(10,3) / C(27,3) = 120/2925 = 8/195

so prob(no blue eyes) = 1 - 8/195 = 187/195

The the odds in favour of all being blue eyes = (8/195) : 187/195
= 8 : 187

Note, do not confuse ODDS with probability, they are not the same

How do you think you can quickly answer part b) ?

Answer 2

C(13,3),C(27,3) = 22/225
1-22/225=203/225
odds=22/225 / 203/225 =22/203

Answer 3

is it correct?

Answer 4

Where did you get 13 from?
How many of the 27 have blue eyes?

Answer 5

Because 10 have blue and 3 have green

Answer 6

So? Where does the green eyes part come in??
Read your question.

Answer 7

Of 27 students in a class, 10 have blue eyes , 14 have brown eyes , and
3 have green eyes. write now I'm just confused.

Answer 8

right*

Answer 9

a.what are the odds of all three have blue eyes?
b.what are the odds of NO one has brown eyes ?

No green eyes!
I am done here

Answer 10

so how do you solve B?

Answer 11

I really don't know how to answer it, you said do you "think" you can solve it and I tried.