8. Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?
a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.
The way I did it is
no of ways 3 blacks can be chosen out of 4 is 4 C 3
Total ways = 10 C 3
answer = 4c3/10c3
another way to solve as the solution proposed -
The probability for the first one to be black is: 4/(4+6) = 2/5.
The probability for the second one to be black is: 3/(3+6) = 1/3.
The probability for the third one to be black is: 2/(2+6) = 1/4.
The probability for all three events is (2/5) x (1/3) x (1/4) = 1/30.
Are both the methods correct or just the second one.
In complex problems how to decide whether to choose the first or second?
Please help
a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.
The way I did it is
no of ways 3 blacks can be chosen out of 4 is 4 C 3
Total ways = 10 C 3
answer = 4c3/10c3
another way to solve as the solution proposed -
The probability for the first one to be black is: 4/(4+6) = 2/5.
The probability for the second one to be black is: 3/(3+6) = 1/3.
The probability for the third one to be black is: 2/(2+6) = 1/4.
The probability for all three events is (2/5) x (1/3) x (1/4) = 1/30.
Are both the methods correct or just the second one.
In complex problems how to decide whether to choose the first or second?
Please help
