Answer:
0.2239
Step-by-step explanation:
Given that you have two hats. In one hat are balls numbered 1 through 15. In the other hat are balls numbered 16 through 25. I first choose a hat, then from that hat, I choose 3 balls, without replacing the balls between selections.
Either I or II hat can be selected.
Prob of selecting any one hat = 0.5
After selecting I hat, selecting 3 balls can be done in 15C3 ways and from II hat it can be done in 10C3 ways.
total no of ways of selecting 3 balls = (10+15)C3 = 2300
Probability required [tex]=\frac{0.5*15C3+0.5*10C3}{2300} \\=\frac{455+60}{2300} \\=0.2239[/tex]
Answer:
3450 ways
Step-by-step explanation:
Different ways of picking 3 balls from the 1st hat:
1st ball: 15 choices (because there are 15 balls)
2nd ball: 14 choices (because one ball is taken away)
3rd ball: 13 choices (because two ball is taken away)
15*14*13=2730 different possible orders of picking 3 balls from the 1st hat
Different ways of picking 3 balls from the 2nd hat:
1st ball: 10 possibilities (because there are 10 balls)
2nd ball: 9 possibilities
3rd ball: 8 possibilities
10*9*8=720 different possible orders of picking 3 balls from the 2nd hat
720+2730=3450
