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

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

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

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.

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)

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

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 lollipopsjellybeans, 3 jellybeanslollipops, and 4 truffles...[the rest remains the same]"

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