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rxs0005
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Pardon my ignorance here. I tend to struggle with probability.
I have a basic doubt common to these type of questions.

The question says the balls were drawn randomly. How do I know whether both the balls were drawn at once or one after the other?
in the first scenario the total P would be 12c2 whereas in the second case it will be 12*11.

How do i tackle this? is there something im missing out on?
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rohu27
Pardon my ignorance here. I tend to struggle with probability.
I have a basic doubt common to these type of questions.

The question says the balls were drawn randomly. How do I know whether both the balls were drawn at once or one after the other?
in the first scenario the total P would be 12c2 whereas in the second case it will be 12*11.

How do i tackle this? is there something im missing out on?

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

Check Ian Stewart's and my posts here: probability-of-simultaneous-events-veritas-vs-mgat-105994-20.html?hilit=simultaneously#p871809

Also note that though 12C2=6*11 is half of 12*11 but the nominator in combinatorial approach is also half of that in probability approach thus they give the same result (check my post above with both approaches giving the same answer).
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thanks a lot Bunnel!! The post clears it all.

Bunuel
rohu27
Pardon my ignorance here. I tend to struggle with probability.
I have a basic doubt common to these type of questions.

The question says the balls were drawn randomly. How do I know whether both the balls were drawn at once or one after the other?
in the first scenario the total P would be 12c2 whereas in the second case it will be 12*11.

How do i tackle this? is there something im missing out on?

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

Check Ian Stewart's and my posts here: probability-of-simultaneous-events-veritas-vs-mgat-105994-20.html?hilit=simultaneously#p871809

Also note that though 12C2=6*11 is half of 12*11 but the nominator in combinatorial approach is also half of that in probability approach thus they give the same result (check my post above with both approaches giving the same answer).
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rxs0005
A jar has 6 green 4 red and 2 blue marbles what is the probability of drawing 2 marbles randomly and they will be of different colors

1/4

1/2

2/3

7/9

8/15



is this approach okay ?

1)1st draw green+2nd draw not green = 6/12*6/11=3/11

2)1st draw red+2nd draw not red =
4/12*8/11=8/33

3)1st draw blue+2nd draw not blue =
2/12*10/11=5/33

now total favorable = 3/11+8/33+5/33
=22/33
=2/3

can some expert look into these approach and verify if it's a valid way to do this as I got the desired answer using this ?? thanks

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