Problem Solving

blackjack wrote:Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jellybeans and 4 truffles. Jack takes one piece of candy out of the bag at random. If he picks a jellybean, he chooses one additional piece of candy and then stops. If he picks any non-jellybean candy, he stops picking immidietely. After jack picks his candy, renee will pick a piece of candy. what is the probability that Renee picks a Jellybean?

Its a PS from Manhattan Word Translations and no options are given. It has just 1 correct answer. [spoiler]3/10[/spoiler]

For some reason I am neither able to break down the question into a simpler form nor able to grasp the textbook explanation. Help?

The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick.

cbaum wrote:

The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick.

Sorry, maybe I'm misunderstanding, but doesn't this imply replacement? And in the problem, it notes that Renee is picking her candy AFTER Jack has already chosen (I assume that he will not be returning to the bag). Could someone please elaborate on why the probability is the same as if Renee picked first instead of some blend of the potential probabilities after Jack has picked? Thanks.

blackjack wrote:Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jellybeans and 4 truffles. Jack takes one piece of candy out of the bag at random. If he picks a jellybean, he chooses one additional piece of candy and then stops. If he picks any non-jellybean candy, he stops picking immidietely. After jack picks his candy, renee will pick a piece of candy. what is the probability that Renee picks a Jellybean?

Its a PS from Manhattan Word Translations and no options are given. It has just 1 correct answer. [spoiler]3/10[/spoiler]

For some reason I am neither able to break down the question into a simpler form nor able to grasp the textbook explanation. Help?

GMATGuruNY wrote:

cbaum wrote:

The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick.

Sorry, maybe I'm misunderstanding, but doesn't this imply replacement? And in the problem, it notes that Renee is picking her candy AFTER Jack has already chosen (I assume that he will not be returning to the bag). Could someone please elaborate on why the probability is the same as if Renee picked first instead of some blend of the potential probabilities after Jack has picked? Thanks.

No replacement is necessary.

Given 3 red marbles and 5 blue marbles:

P(R) on the first pick = 3/8.

P(R) on the second pick:
P(RR) = 3/8 * 2/7 = 6/56.
P(BR) = 5/8 * 3/7 = 15/56.
Adding the probabilities, we get:
P(R) on the second pick = 6/56 + 15/56 = 21/56 = 3/8.

P(R) on the third pick:
P(RRR) = 3/8 * 2/7 * 1/6 = 1/56.
P(RBR) = 3/8 * 5/7 * 2/6 = 5/56.
P(BRR) = same as P(RBR) = 5/56.
P(BBR) = 5/8 * 4/7 * 3/6 = 10/56.
Adding the probabilities, we get:
P(R) on the third pick = 1/56 + 5/56 + 5/56 + 10/56 = 21/56 = 3/8.

P(R) on the Nth pick = P(R) on the first pick = 3/8.

mjmehta81 wrote:

GMATGuruNY wrote:

cbaum wrote:

The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick.

Sorry, maybe I'm misunderstanding, but doesn't this imply replacement? And in the problem, it notes that Renee is picking her candy AFTER Jack has already chosen (I assume that he will not be returning to the bag). Could someone please elaborate on why the probability is the same as if Renee picked first instead of some blend of the potential probabilities after Jack has picked? Thanks.

No replacement is necessary.

Given 3 red marbles and 5 blue marbles:

P(R) on the first pick = 3/8.

P(R) on the second pick:
P(RR) = 3/8 * 2/7 = 6/56.
P(BR) = 5/8 * 3/7 = 15/56.
Adding the probabilities, we get:
P(R) on the second pick = 6/56 + 15/56 = 21/56 = 3/8.

P(R) on the third pick:
P(RRR) = 3/8 * 2/7 * 1/6 = 1/56.
P(RBR) = 3/8 * 5/7 * 2/6 = 5/56.
P(BRR) = same as P(RBR) = 5/56.
P(BBR) = 5/8 * 4/7 * 3/6 = 10/56.
Adding the probabilities, we get:
P(R) on the third pick = 1/56 + 5/56 + 5/56 + 10/56 = 21/56 = 3/8.

P(R) on the Nth pick = P(R) on the first pick = 3/8.

Dear Mitch,

I have kept in my mind your thumb rule "The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick." But recently going through explanation of 'Domino Effect' in Manhattan Probability Strategy lesson, following example is given:

In a box with 10 blocks, 3 of which are red, what is probability of picking out a red block on each of your first two tries? Consider no replacement.

Solution :
P(R) on first pick - 3/10

P(R) on second pick - 2/9 (DOMINO EFFECT - If red block is chosen on first pick, then the number of blocks in box reduced to 9 from 10. Additionally, the number of red blocks now decreased to 2 from 3. So P(R) on second pick is different from P(R) on first pick. )

Therefore, P(R) on both picks is 3/10 x 2/9 = 6/90= 1/15. Ans.

But as per your explained thumb rule, for above example P(R) on 2nd pick = P(R) on 1st pick = 3/10
therefore Ans = 3/10 x 3/10 = 9/100. But this is clearly mentioned as 'WRONG SOLUTION' in Manhattan Book.

Please help to understand why the thumb rule cannot be applicable to above example and explanation of 'Domino Effect'.

Thanks,

Mrudang Mehta

krusta80 wrote: The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick. <--- ONLY APPLIES TO SELECTION WITH REPLACEMENT

In a box with 10 blocks, 3 of which are red, what is probability of picking out a red block on each of your first two tries? Consider no replacement.

Solution :
P(R) on first pick - 3/10

P(R) on second pick - 2/9 (DOMINO EFFECT - If red block is chosen on first pick, then the number of blocks in box reduced to 9 from 10. Additionally, the number of red blocks now decreased to 2 from 3. So P(R) on second pick is different from P(R) on first pick. )

Therefore, P(R) on both picks is 3/10 x 2/9 = 6/90= 1/15.

The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick.

GMATGuruNY wrote:

krusta80 wrote: The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick. <--- ONLY APPLIES TO SELECTION WITH REPLACEMENT

The rule does not require that selections be made without replacement.
Please see my post above, in which my calculations assume no replacement.
Given 3 red marbles and 5 blue marbles, P(R) on any given pick (with or without replacement) = 3/8.

In a box with 10 blocks, 3 of which are red, what is probability of picking out a red block on each of your first two tries? Consider no replacement.

Solution :
P(R) on first pick - 3/10

P(R) on second pick - 2/9 (DOMINO EFFECT - If red block is chosen on first pick, then the number of blocks in box reduced to 9 from 10. Additionally, the number of red blocks now decreased to 2 from 3. So P(R) on second pick is different from P(R) on first pick. )

Therefore, P(R) on both picks is 3/10 x 2/9 = 6/90= 1/15.

The solution above determines the probability that BOTH blocks are red.
Thus, there are TWO events that must happen: red on the first pick AND red on the second pick.
As the solution shows, the probability that BOTH of these events happen = 1/15.

The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick.

This rule applies to ONE EVENT ONLY: the probability of selecting X on any given pick.
It does not apply to the block problem, which asks for the probability of TWO EVENTS: red on the first pick AND red on the second pick.
While the probability of selecting red on BOTH picks is 1/15, the probability of selecting red on the SECOND pick -- with no restrictions on the first pick -- is greater.
The reason is that there is MORE THAN ONE WAY to get red on the second pick:

Case 1: First block is not red, second block is red
P(first block is not red) = 7/10. (Of the 10 blocks, 7 are not red.)
P(second block is red) = 3/9. (Of the 9 remaining blocks, 3 are red).
Since we want both of these events to happen, we multiply the fractions:
7/10 * 3/9 = 7/30.

Case 2: Both blocks are red
As we saw above, P(both blocks are red) = 1/15.

Since each case is a way to get red on the second pick, we add the fractions:
P(red on the second pick) = 7/30 + 1/15 = 9/30 = 3/10.

Notice that the calculations for each case take into account the "domino" effect.

Notice also that -- when all of the different ways to get red on the second pick are considered -- the probability of selecting red on the second pick, WITHOUT replacement, is equal to the probability of selecting red on the FIRST pick: 3/10.