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 .
what are the odds of all three have blue 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
Not very nice to pass on somebody's else solution as your own !
I agree with Reiny.
MEF/sara/hola — not too smart to keep switching names.
Connexus student — a plagiarizer
https://www.jiskha.com/questions/1823069/of-27-students-in-a-class-10-have-blue-eyes-14-have-brown-eyes-and-3-have-green-eyes
To find the odds of all three students having blue eyes, we need to determine the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes: When three students are chosen at random from a class of 27, there are a total of 27C3 possible outcomes. This is because we are selecting a combination of 3 students from a group of 27.
The formula for combination is nCk = n! / (k!(n-k)!), where n is the total number of students and k is the number of students we are selecting (in this case, 3).
So, 27C3 = 27! / (3!(27-3)!) = 27! / (3! * 24!) = (27 * 26 * 25) / (3 * 2 * 1) = 2925.
Number of favorable outcomes: We want all three students to have blue eyes. Since there are 10 students with blue eyes, we need to select all three students from this group.
This can be calculated as 10C3 = 10! / (3!(10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.
So, the number of favorable outcomes is 120.
Now, we can calculate the odds by dividing the number of favorable outcomes by the total number of possible outcomes:
Odds = Number of favorable outcomes / Total number of possible outcomes
= 120 / 2925
≈ 0.041 or 4.1%
Therefore, the odds of all three chosen students having blue eyes is approximately 4.1%.