In a 3-ball lottery, three numbered balls are released from a chamber. the number of even-numbered balls in the chamber is twice the number of odd numbered balls. what is the probability that at least 2 of the 3 released balls will be odd? assume that there are 12 in the chamber and balls are released without replacement.
I have obviously tried this, but there is not point in me showing you my wrong working out as it is quite long. I hope someone can help me here...
answer is 4/27
I have obviously tried this, but there is not point in me showing you my wrong working out as it is quite long. I hope someone can help me here...
answer is 4/27
-
there are 4 odd #s & 8 even #s, total 12
we want {2 odd, 1 even} or {3 odd}
use combinations
Pr = (4c2*8c1 + 4c3)/12c3 = 52/220 = 13/55 <-------
the given ans is wrong. let's doublecheck with direct probabilities
3* 4/12 *3/11 *8/10 + 4/12 *3/11 *2/10 = 13/55 <-------
we want {2 odd, 1 even} or {3 odd}
use combinations
Pr = (4c2*8c1 + 4c3)/12c3 = 52/220 = 13/55 <-------
the given ans is wrong. let's doublecheck with direct probabilities
3* 4/12 *3/11 *8/10 + 4/12 *3/11 *2/10 = 13/55 <-------
-
im bad at math