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

Question

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

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

CareerGeek · Accepted Answer

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

Lampard42 · Answer

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

Archit3110 · Answer

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

stne · Answer

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

philipssonicare · Answer

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.

exc4libur · Answer

Assuming that Renee and Jack do not replace the candies taken out;
Jack picks first, if a Jelly then pick one more time; Renee picks next;
c=1, lol=2, j=3, t=4; total=10; not=7/10, j=3/10;

P(R=j) cases:
NJ: 7/10*3/9 = 21/90
JNJ: 3/10*7/9*2/8 = 42/90*8
JJJ: 3/10*2/9*1/8 = 6/90*8

P(R=j) total:
21/90+(42+6)/90*8=(8*21+48)/90*8=(21+6)/90=27/90=3/10

Ans (C)

Regor60 · Answer

Since there is no information as to which candy Jack picks, the probability of Renne's picking a jellybean is the same as Jack's, 3/10.

In other words, the narrative provided is irrelevant.

Posted from my mobile device

ThatDudeKnows · Answer

Regor60

What if we change the prompt to the following...does your approach still apply?

"Renne has a bag of candy. The bag has 1 candy bar, 2  lollipops  jellybeans , 3  jellybeans  lollipops , and 4 truffles... "

Regor60 · Answer

My approach would suggest 1/5.

We can test the long way:

I. J picks a non jelly so prob =

8/10*2/9 = 16/90

II) J picks a Jelly in First go and non jelly in Second go =

2/10*8/9*1/8= 2/90

III) J Picks Jelly in first and second go :

2/10*1/9*1*0 = 0

Total probability: 18/90 =

1/5

webarebears · Answer

Bunuel  is this true? Can you help confirming this one? Thank you

skasmani · Answer

That’s what i thought hence 3/10

Bunuel · Answer

Yes. Check similar questions below: https://gmatclub.com/forum/a-box-contai ... 90272.html https://gmatclub.com/forum/a-bag-contai ... 00023.html https://gmatclub.com/forum/each-of-four ... 01553.html https://gmatclub.com/forum/if-40-people ... 97015.html https://gmatclub.com/forum/a-bag-contai ... 00023.html https://gmatclub.com/forum/nine-cards-n ... 47082.html Hope this helps.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82871 --- Harsha Independent GMAT Coach | My website: gmatanchor.com GMAT-Club-Forum-594epfe8.png

bumpbot · Answer

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

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