1. ## disc drawn

a bag contains 4 red, 3 blue,5 white markers. three markers are drawn. whats the probability all three are different colour.?

so there must be one red,one blue,one white drawn.

say a red is drawn first so 4/12 than the probability of the next one being drawn not being red is 8/11 but i can not work out the third one becuase for the second disc drawn i do not know if it is blue or white so i am confused

2. ## Re: disc drawn

There are 3!= 6 possible cases for the 6 different orders:
Red, Blue, White
Red, White, Blue
Blue, Red, White
Blue, White, Red
White, Blue, Red
White, Red, Blue

Fortunately, it is easy to see that these cases all involve basically the same calculation.
Red, Blue, White: There are a total of 12 markers, 4 of them red. The probability the first drawn is red is 4/12. There are then 11 markers left, 3 of them blue. The probability the next drawn is blue is 3/11. There are then 10 markers left, 5 of them white. The probability the last drawn is white is 5/10. The probability the three markers drawn are "Red, Blue, White", in that order, is (4/12)(3/11)(5/10)= 1/22.

Red, Blue, White: There are a total of 12 markers, 4 of them red. The probability the first drawn is red is 4/12= 1/3. There are then 11 markers left, 5 of them white. The probability the next drawn is white is 5/11. There are then 10 markers left, 3 of them blue. The probability the last drawn is white is 3/10. The probability the three markers drawn are "Red, White, Blue", in that order, is (1/3)(5/11)(3/10)= 1/22 also. The fractions are different but the numerators and denominators are the same and that this will be true for all orders of "Red, Blue, and White".

So there are 6 possible cases, all with probability 1/22. The probability any one of those will happen is 6(1/22)= 3/11.

3. ## Re: disc drawn

thank you. laying it out the way you did makes it clear