Question on Probability

[Vemuri]

A bag contains 6 marbles., 3 blue & 3 green. What is the probability of blindly reaching into the bag and pulling out 2 green marbles?

I am confused on how to intrepret this question.

Simultaneously
If the person is randomly taking out 2 green marbles at the same time, then the probability is 3c2/6 ==> 3/6 ==> 1/2

Sequentially
But, if he is taking out 1 green marbles first, the probability for which is 3/6 & then takes out another green marble (2/5), then the total probability is = 3/6*2/5 ==> 1/5

How should I be intrepreting the question & are the 2 solutions above correct in their respective way?

Appreciate your responses.

[mals24]

Your solutions are correct in their respective ways.

But then this question should be interpreted according to the first method that you've done because the question says that when you are blindly pulling out the marbles you are picking 2 marbles in the first attempt it self.

If there were 2 separate attempts the question would have clearly stated that what is the probability of picking 1 green marble and another green marble without/with replacing the first ball.
So if there were 2 separate attempts made, you would be given data stating whether the first marble picked was replaced into the bag or not. (Since your answer would be different for both the cases replaced/not replaced)

[Vemuri]

Thanks mals24.

[Ian Stewart]

A bag contains 6 marbles., 3 blue & 3 green. What is the probability of blindly reaching into the bag and pulling out 2 green marbles?

I am confused on how to intrepret this question.

Simultaneously
If the person is randomly taking out 2 green marbles at the same time, then the probability is 3c2/6 ==> 3/6 ==> 1/2

Sequentially
But, if he is taking out 1 green marbles first, the probability for which is 3/6 & then takes out another green marble (2/5), then the total probability is = 3/6*2/5 ==> 1/5

How should I be intrepreting the question & are the 2 solutions above correct in their respective way?

Appreciate your responses.

Mathematically, picking 'two marbles at once' is the same as 'picking one marble, then picking another, without putting the first marble back'. You can solve this problem taking either perspective, as long as we count the numerator and denominator of the probability in the same way:

- Picking both marbles at the same time, we can pick two green marbles from the three available in 3C2 ways. From the six marbles, we can pick two in 6C2 ways. So the probability of picking two green marbles is 3C2/6C2 = 3/15 = 1/5 (you didn't get the correct answer here, because your denominator wasn't correct).

- Picking one at a time, there is a 3/6 chance the first marble is green, and then a 2/5 chance the second is also green. (3/6)*(2/5) = 1/5.

I normally find the 'one at a time' approach easier, personally.

[cramya]

Great explanation and thanks Ian.

Regards,
CR

[mals24]

- Picking both marbles at the same time, we can pick two green marbles from the three available in 3C2 ways. From the six marbles, we can pick two in 6C2 ways. So the probability of picking two green marbles is 3C2/6C2 = 3/15 = 1/5 (you didn't get the correct answer here, because your denominator wasn't correct).

Sorry might sound like a weird question but why is the total number of possible outcomes (denominator) in the above case 6C2 and not just 6.

Ian can you elaborate a little bit further on this point.

[kapsii]

total possible outcomes is the number of ways you can choose 2 balls out of the six given. Thus 6C2.

6 => total number of balls you have initially to choose from...

I guess the confusion is stemming from the statement that if balls are picked up one at a time probability is (3/6)*(2/5)... You should read that as (3C1/6C1)*(2C1/5C1) cause after all by definition probability is the ratio of number of favorable outcomes to total possible outcomes.

[Vemuri]

Mathematically, picking 'two marbles at once' is the same as 'picking one marble, then picking another, without putting the first marble back'. You can solve this problem taking either perspective, as long as we count the numerator and denominator of the probability in the same way:

- Picking both marbles at the same time, we can pick two green marbles from the three available in 3C2 ways. From the six marbles, we can pick two in 6C2 ways. So the probability of picking two green marbles is 3C2/6C2 = 3/15 = 1/5 (you didn't get the correct answer here, because your denominator wasn't correct).

- Picking one at a time, there is a 3/6 chance the first marble is green, and then a 2/5 chance the second is also green. (3/6)*(2/5) = 1/5.

I normally find the 'one at a time' approach easier, personally.

Wonderful explanation Ian. Thanks for correcting my understanding.