a bag contains 4 red, 6 green and 4 blue balls. what is the probability that none of the balls drawn is blue, when two balls are drawn from the bag?
The total number of balls in the bag is 4 + 6 + 4 = 14.
The number of ways to select 2 balls from the 14 is given by the combination formula:
nCr = n! / r!(n - r)!
where n is the total number of balls and r is the number of balls being drawn.
So, the number of ways to select 2 balls from 14 is:
14C2 = 14! / 2!(14 - 2)!
14C2 = 91
Now, we need to find the probability that none of the balls drawn is blue, which means we have to select both the balls from the 4 red and 6 green balls. The number of ways to select 2 balls from the 4 red and 6 green balls is:
10C2 = 10! / 2!(10 - 2)!
10C2 = 45
Therefore, the probability of selecting 2 balls such that none of them is blue is:
45 / 91 = 0.4945 (rounded to four decimal places)
So, the probability that none of the balls drawn is blue, when two balls are drawn from the bag, is approximately 0.4945.
a bag contains 4 red, 6 green and 4 blue balls. what is the probability that none of the balls drawn is blue, when two balls are drawn from the bag? @aibot