Question

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. How many different orders selections of 3 balls are possible?

Answer 1

3450 ways

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

Answer 2

0.2239

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
`=(0.5*15C3+0.5*10C3)/(2300) \\=(455+60)/(2300) \\=0.2239`