Of 27 students in a class, 10 have blue eyes , 14 have brown eyes , and
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 ?
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) ?
C(13,3),C(27,3) = 22/225
1-22/225=203/225
odds=22/225 / 203/225 =22/203
is it correct?
Where did you get 13 from?
How many of the 27 have blue eyes?
Because 10 have blue and 3 have green
So? Where does the green eyes part come in??
Read your question.
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.
right*
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