What Is the Probability of Drawing Specific Ball Colors from a Bag?

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SUMMARY

The discussion centers on calculating the probability of drawing specific colored balls from a bag containing 5 red, 4 blue, and 3 green balls. When drawing two balls sequentially without replacement, the probability of both balls being red is determined by the formula P(Red 1st) * P(Red 2nd | Red 1st), resulting in a probability of 15/121. Conversely, the probability of drawing at least one green ball, given that the first ball drawn is not green, requires calculating the complementary probability of drawing only red and blue balls, yielding a probability of 8/11.

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Hi guys, I have a maths question about Probability, hope somebody can help me out, thank you very much.

There are 5 red, 4 blue and 3 green balls in a bag. Two balls are taken from the bag, one after the other. If the first ball is not replaced, and it was not green. What is the probability that:

a) both balls are red?
b) at least one ball is green?
 
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raindoe said:
Hi guys, I have a maths question about Probability, hope somebody can help me out, thank you very much.

There are 5 red, 4 blue and 3 green balls in a bag. Two balls are taken from the bag, one after the other. If the first ball is not replaced, and it was not green. What is the probability that:

a) both balls are red?
b) at least one ball is green?
a) There are a total of 12 balls in the bag, 3 of them red. What is the probability the ball drawn is red? If the first ball is red there are 11 balls left, 2 of them red. What is the probability the second ball is also red? So what is the probability they are both red?

b) You are told that the first ball drawn is "not green". "At least one ball is green" requires that the second ball also be green. There are 11 balls left. What is the probability the second ball is "green"?
 
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