roshpat wrote:
almostfamous wrote:
Your logic is correct.
Regardless of the number of marbles to be selected, the underlying logic for this type of questions hinges on the same one as the one for simultaneous picking.
In your example:
Red: 3
White: 2
Black: 5
Total: 10
As we are picking the marbles "simultaneously" (i.e. without replacement), the probability is:
3/10 x 2/9 x 5/8 x 4/7 = 1/42
Hope this helps!
Sorry, I am going to raise one more quick question with you. I am just going over the
MGMAT questions after going through Veritas and am a bit muddled.
There is an example in the
MGMAT book -
"A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?"
In this case, if we are to apply the Veritas logic, that the order does not matter, simply:
7/16* 6/15*4/14 = 1/24
MGMAT goes on to say that there are 3! ways to calculate this and multiplies 1/24 with 3! to get 1/4.
Just wanted to check - this is an overestimate right and the correct answer is 1/24, isnt it?
Would really appreciate a clarification on this point.
Thanks very much.
No. The answer to the question above is 1/4.
The machine dispenses the 3 gumballs, one after another.
It could do it in 6 different ways: BGR, GRB, BRG, etc
Explanation: The 'Veritas Method' (actually it is simultaneous events method) is used when it is mentioned 'three balls are dispensed
simultaneously'. Here the questions says, 'three balls are dispensed. What is the probability that one is red, one green and one blue?' It does not say that the three balls are dispensed simultaneously.
There is a difference.
Case 1: There is one red, one blue and one green ball. I pick two balls. What is the probability that one is red and one blue?
Here I pick two balls. It doesn't say I pick them simultaneously. I can pick them in 6 ways:
1 R, 1 B
1B, 1 R
1 R, 1G
1G, 1R
1B, 1G
1G, 1B
Probability of each is 1/6. Probability of 1 red and 1 blue is 2*1/6 = 1/3 since there are 2 such cases.
Case 2: There is one red, one blue and one green ball. I pick two balls
simultaneously. What is the probability that one is red and one blue?
Now the order does not matter. The probability of 1/3 we got above is divided by 2 because 1/3 includes 2 cases: 1R, 1B and 1B, 1R.
So the probability here is 1/6. Same as the probabilty of 1 Red and then 1 Blue. Also same as probability of 1 Blue and then 1 Red.
It is counter intuitive. But mathematicians have proved it conclusively.