# in a carnival there are 95 red balls and 30 blue balls. you get to pick 3 random balls without putting them back,?

- and if you get a blue ball you get a prize.

what is the probability of getting a blue ball? how do you solve this?

### 4 Answers

- MorningfoxLv 71 year ago
The probability of getting all red balls is (95/125) * (94/124) * ( 93/123) = 43.56%. So the prob. of getting a blue ball is 100% - 43.56% = 56.44%.

- A.J.Lv 71 year ago
Easiest way is assuming one or more blue balls get a prize since nothing stated about extra blue ones.

All red is the complement to one or more blue.

3 red

95+30=125 total

95/125 first pick

94/124 second pick

93/123 third pick

(95 x 94 x 93) / (125 x 124 x 123) is probability of 3 red

1- ((95 x 94 x 93) / (125 x 124 x 123)) is probability of 1 or more blue

1- 830490/1906500

1- 0.43 56097 56097...

0.56439

- 1 year ago
Again, examine your possibilities:

0 blue balls

1 blue ball

2 blue balls

3 blue balls

The only way you lose is if you get no blue balls

95 red balls + 30 blue balls = 125 balls

95/125 chance of pulling a red ball on the first try

94/124 chance of pulling a red ball on the 2nd try

93/123 chance of pulling a red ball on the 3rd try

The chances of pulling 3 red balls is:

(95/125) * (94/124) * (93/123) =>

(19/25) * (47/62) * (31/41) =>

19 * 47 * 31 / (25 * 62 * 41) =>

19 * (39 + 8) * (39 - 8) / (25 * 2 * 31 * 41) =>

19 * (39^2 - 8^2) / (50 * (36 - 5) * (36 + 5)) =>

19 * ((40 - 1)^2 - 64) / (50 * (36^2 - 5^2)) =>

19 * (1600 - 80 + 1 - 64) / (50 * (1296 - 25)) =>

19 * (1600 - 143) / (50 * 1271) =>

19 * 1457 / (50 * (1270 + 1)) =>

(20 - 1) * 1457 / (50 * 2 * 635 + 50) =>

(29140 - 1457) / (100 * 635 + 50) =>

(28140 - 457) / (63500 + 50) =>

(27740 - 57) / 63550 =>

(27690 - 7) / 63550 =>

27683 / 63550

1 - (27683 / 63550) =>

(63550 - 27683) / 63550 =>

(36550 - 683) / 63550 =>

(39950 - 83) / 63550 =>

(39870 - 3) / 63550 =>

39867 / 63550