almostfamous wrote:
Hi,
Background: I am studying probability, one of my weakest areas, using both
MGMAT and
Veritas books.
The probability of a simultaneous events (picking two balls without replacement) is explained DIFFERENTLY in those books. I am turning to wise Clubbers to point out which is right.
Question: there are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles? (
MGMAT Word Translation, p93, Q9)
The correct answer, according to
MGMAT is 2/9. You can pick first a white marble and then a blue marble which is 1/5 x 5/9 = 1/9. You can also pick the blue first and then white: 1/2 x 2/9 = 1/9.
MGMAT then urges to add up both probabilities and get 2/9 as the correct answer.
In the
Veritas Combinatorics, the correct answer is 1/9 i.e. in simultaneous events, you do not add up the both cases.
My hunch is that
Veritas is correct. I think this, because the probability of a simultaneous event (or event without replacement) is ALWAYS THE SAME regardless of the order (i.e. if you pick white or blue marble first). I think that if you add up, you are overestimating the probability of the event.
As said, I am weak at probability, and want to hear what other, wiser Clubbers say! Please let me know which book got it right...!
MGMAT answer is correct.
There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?P(WB)=2*2/10*5/9=2/9, we multiply by 2 as the scenario WB can occur in two ways: first ball is white and the second is blue OR first ball is blue and the second is white.
VeritasPrepKarishma wrote:
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.
I don't think the above is correct.
Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.So the answer for both questions below is 1/3.
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?
P=2*1/3*1/2=1/3.
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?
You can do the same as above or consider this: P=# of favorable outcomes/total # of outcomes --> there are total of 3 outcomes possible for 2 balls: (Red, Blue), (Red, Green), and (Blue, Green) --> # of favorable outcomes is 1: (Red, Blue) --> P=1/3 (or \(P=\frac{C^2_2}{C^2_3}=\frac{1}{3}\)).
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