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# Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jell

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Math Expert
Joined: 02 Sep 2009
Posts: 57193
Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jell  [#permalink]

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17 Jun 2019, 01:35
10
00:00

Difficulty:

55% (hard)

Question Stats:

65% (03:03) correct 35% (03:10) wrong based on 60 sessions

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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 immediately. After jack picks his candy, renee will pick a piece of candy. what is the probability that Renee picks a Jellybean?

A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 7/10

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Director
Joined: 20 Jul 2017
Posts: 635
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jell  [#permalink]

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17 Jun 2019, 02:37
2
Bunuel 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 immediately. After jack picks his candy, renee will pick a piece of candy. what is the probability that Renee picks a Jellybean?

A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 7/10

2 cases are possible:
Case 1: Jack picks a jellybean(JB) and then picks another JB or a non JB and stops
Case 2: Jack picks a non JB and stops

Probability of Renee picks a JB = Prob(Case 1)*Prob(Renee picking a JB) + Prob(Case 2)*Prob(Renee picking a JB)
= 3/10*2/9*1/8 (Jack picking JB & JB & Renee picking a JB) + 3/10*7/9*2/8 (Jack picking a JB & Non JB & Renee picking a JB) + 7/10*3/9 (Jack picking a JB & Renee picking a JB)
= 1/120 + 7/120 + 7/30
= 8/120 + 28/120
= 36/120
= 3/10

Pls Hit Kudos if you like the solution
Director
Joined: 22 Nov 2018
Posts: 534
Location: India
GMAT 1: 640 Q45 V35
GMAT 2: 660 Q48 V33
Re: Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jell  [#permalink]

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17 Jun 2019, 03:22
1
Bunuel 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 immediately. After jack picks his candy, renee will pick a piece of candy. what is the probability that Renee picks a Jellybean?

A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 7/10

Three Possibility exists

I) J picks a non jelly so prob = 7/10*3/9 = 7/30

II) J picks a Jelly in First go and other in Second go : 3/10*7/9*2/8= 7/120

III) J Picks Jelly in first and second go : 3/10*2/9*1/8 = 1/120

So Renne picking a jelly is 28/120+1/120+7/120 = 36/120; 3/10 IMO C
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Re: Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jell  [#permalink]

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17 Jun 2019, 11:37
2
Bunuel 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 immediately. After jack picks his candy, renee will pick a piece of candy. what is the probability that Renee picks a Jellybean?

A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 7/10

understanding the wording took me >2 mins IMO its a very confusing question ; main objective of question is to find probability that Renee picks a Jellybean in all possible cases with Jack being part in all draws...

so we have three cases;
total candies = 10
case 1; Jack takes 1 jelly bean and 1 additional non JB and then Renee takes JB ; 3/10 * 7/9*2/8 ; 7/120
case 2: jack takes 2 jelly bean and 1 jelly bean by Reenee; 3/10 * 2/9 * 1/8 = 1/20
case 3; Jack takes 1 non JB and 1 JB by Reenee; 7/10 * 3/9= 7/30

total P = 7/120+1/20+7/30 ; 36/120 ; 3/10
IMO C
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Director
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Re: Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jell  [#permalink]

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08 Aug 2019, 04:17
Bunuel 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 immediately. After jack picks his candy, renee will pick a piece of candy. what is the probability that Renee picks a Jellybean?

A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 7/10

Although most of the people have got the answer correct, there is one particular step that needs to be corrected.

Case: Jack gets no Jelly Bean and Renne gets a Jelly bean
7/10*3/9= 7/30

Case : Jack gets a single JB then non JB and then Renne gets a JB
3/10*7/9*2/8=7/120

Case :When Jack picks two Jelly beans in a row: ( This step needs special focus )
3/10*2/9*7/8*1/7 =1/120

Jack gets a Jelly bean 3/10
Jack Gets a Jelly bean second time 2/9
Jack gets a non Jelly Bean=7/8
Renne gets the single remaining Jelly bean= 1/7

Note: as long as Jack gets a Jelly bean he has to pick another candy till he gets a Non Jelly Bean.

Case : Jack gets 3 JB in a row and Renne gets no JB
3/10*2/9*1/8*0/7= 0

So Probability that Renne gets a JB
7/30 + 7/120 +1/120 + 0 = 3/10
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Posts: 299
Re: Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jell  [#permalink]

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12 Aug 2019, 16:46
I understand the probability in the following scenario: If there are 3 jelly beans and 7 other lollies, and two lollies have been removed, what is the chance of the 3rd being a jelly bean?
In the above, it should be 3/10 still.

I solved the OP, but should a restriction like the OP where repeated removals only happen if JB is taken alter the odds? Seems counter-intuitive and solving hasn't brought understanding.
Re: Renne has a bag of candy. The bag has 1 candy bar, 2 lollipops, 3 jell   [#permalink] 12 Aug 2019, 16:46
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