A bag contains 6 black balls,5 red balls and 4 white balls,if 3 are selected at random without replacement determine the probability that (1)all 3 are black (2)all 3 are red (3)2 are black and 1 white (4)at least 1 of each colour is selected.
To determine the probabilities of the given cases, we will use the concept of combinations and probabilities.
First, let's calculate the total number of ways to select 3 balls from a bag of 15 (6 black, 5 red, and 4 white) without replacement:
Total Number of Ways to Select 3 Balls = C(15, 3) = 15! / (3! * (15-3)!) = 455
Now, let's calculate the probabilities for each case:
1) Probability that all 3 balls are black:
To find this probability, we need to determine the number of ways to select 3 black balls (C(6, 3)), and then divide it by the total number of ways to select 3 balls.
Number of Ways to Select 3 Black Balls = C(6, 3) = 6! / (3! * (6-3)!) = 20
P(All 3 are Black) = Number of Ways to Select 3 Black Balls / Total Number of Ways to Select 3 Balls = 20 / 455
2) Probability that all 3 balls are red:
To find this probability, we need to determine the number of ways to select 3 red balls (C(5, 3)), and then divide it by the total number of ways to select 3 balls.
Number of Ways to Select 3 Red Balls = C(5, 3) = 5! / (3! * (5-3)!) = 10
P(All 3 are Red) = Number of Ways to Select 3 Red Balls / Total Number of Ways to Select 3 Balls = 10 / 455
3) Probability that 2 balls are black and 1 is white:
To find this probability, we need to determine the number of ways to select 2 black balls (C(6, 2)) and 1 white ball (C(4, 1)), and then multiply them together. Finally, divide the result by the total number of ways to select 3 balls.
Number of Ways to Select 2 Black Balls = C(6, 2) = 6! / (2! * (6-2)!) = 15
Number of Ways to Select 1 White Ball = C(4, 1) = 4! / (1! * (4-1)!) = 4
Number of Ways to Select 2 Black Balls and 1 White Ball = Number of Ways to Select 2 Black Balls * Number of Ways to Select 1 White Ball = 15 * 4 = 60
P(2 Black and 1 White) = Number of Ways to Select 2 Black Balls and 1 White Ball / Total Number of Ways to Select 3 Balls = 60 / 455
4) Probability that at least 1 of each color is selected:
To find this probability, we will calculate the complement probability of the event that no black ball, no red ball, or no white ball is selected.
No Black: Number of Ways to Select 3 Non-Black Balls = C(9, 3) = 9! / (3! * (9-3)!) = 84
No Red: Number of Ways to Select 3 Non-Red Balls = C(10, 3) = 10! / (3! * (10-3)!) = 120
No White: Number of Ways to Select 3 Non-White Balls = C(11, 3) = 11! / (3! * (11-3)!) = 165
Number of Ways to Not Have All the Colors Selected = Number of Ways to Select 3 Non-Black Balls + Number of Ways to Select 3 Non-Red Balls + Number of Ways to Select 3 Non-White Balls = 84 + 120 + 165 = 369
P(At Least 1 of Each Color is Selected) = 1 - (Number of Ways to Not Have All the Colors Selected / Total Number of Ways to Select 3 Balls) = 1 - 369 / 455
Note: You can simplify the fractions to obtain decimal values as needed.