Question

A bag contains 666 red balls, 444 green balls, and 333 blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?

Answer 1

0.142110316 or 1332/9373

Explanation:

According to the question, A bag contains 666 red balls, 444 green balls, and 333 blue balls. If we choose a ball, then another ball without putting the first one back in the bag and we are now told to find the probability that the first ball will be green and the second will be red?

Before we proceed I'd like to say that what we have here is an enormous amount of balls and who knows the kind of bag that could contain them.

Anyways,we proceed to the real deal.

We Calculate the total number of balls for a start:

Total number of Balls = Red Balls + Green Balls + Blue Balls

Total Balls = 666 + 444 + 333

Total Balls = 1,443

Then we calculate the probability of drawing a green ball on the

first pick:

P(Green) = Green Balls / Total Balls

P(Green) = 444/1443

P(Green) = 4/13

Calculate the probability of drawing a red ball on the

second pick (without replacement):

Total number of Balls decrease by 1, since we do not replace. So Total

Balls = 1,443 - 1 = 1,442

P(Red) = Red Balls / Total Balls

P(Red) = 666/1442

P(Red) = 333/721

Now, we want the probability of Green, Red in that order.

Since each event is independent, we multiply the event

probabilities

P(Green, Red) = P(Green) × P(Red)

P(Green, Red) =4/13 × 333/721

P(Green, Red) = 1332/9373 or 0.142110316

Answer 2

`P(GR)=(2)/(13)`

Explanation:

Number Of Red Balls=6

Number Of Green Balls=4

Number Of blue Balls=3

Total Number Of Balls=6+4+3=13

Notice that the two balls are picked without replacement. In this case, the total number of balls always decreases.

P(first ball will be green and the second will be red)

`=(4)/(13) X (6)/(12) \\=(2)/(13)`