among 10 white balls and 3 black balls drawn one by one without replacement till all black balls are drawn. probability that all black balls are drawn by the 6th draw

Question

a bag contain 10 white and 3 black balls.balls are drawn one by one without replacement till all the black balls are drawn.find the probability that all black balls are drawn by the 6th draw

Answer 1

There are 20 ways of mixing 3 black balls with 3 white:

WWWBBB, WWBWBB, WWBBWB, WWBBBW, WBWWBB,

WBWBWB, WBWBBW, WBBWWB, WBBWBW, WBBBWW,

BWWWBB, BWWBWB, BWWBBW, BWBWWB, BWBWBW,

BWBBWW, BBWWWB, BBWWBW, BBWBWW, BBBWWW.

We can pick any one of these to work out the probability (P), so let's randomly pick two: WBWBWB and BBBWWW.

P(WBWBWB)=10W out of 13 balls (10W+3B)=10/13, 12 balls left (9W+3B); 3B out of 12=3/12, 11 balls left (9W+2B); 9W/11=9/11, 10 left (8W+2B); 2/10, 9 left (8W+1B); 8/9, 8 left (7W+1B); 1/8. P=10/13*3/12*9/11*2/10*8/9*1/8=1/286.

P(BBBWWW)=3B/13=3/13; 2/12; 1/11; 10/10=1; 9/9=1; 8/8=1. P=3/13*2/12*1/11=1/286.

So, as long as the result is three black and three white balls, the probability is 1/286 for any one of the twenty arrangements. If you look at the numerators and denominators, the denominators are 13*12*11*10*9*8, while the numerators are 1*2*3*8*9*10. The order of multiplication doesn't affect the result. Therefore the probability of all three black balls being drawn in 6 draws is 20/286=10/143=0.06993 or about 7%.