Last visit was: 18 Nov 2025, 18:42 It is currently 18 Nov 2025, 18:42
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
805+ Level|   Probability|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,078
 [23]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,078
 [23]
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 20 Jan 2023, 03:26
Kudos
Add Kudos
23
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

46% (02:27) correct 54% (03:11) wrong based on 174 sessions

History

Date
Time
Result
Not Attempted Yet
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 bees. For every bee received, there is a 50% chance that it is a male and a 50% chance that it is a female. If a jar has at least 1 male, but more females than males, the honey farm keeps the jar; otherwise the jar is returned. If:

    a = the probability that Jar 1 is kept.

    b = the probability that Jar 2 is kept.

What is the value of a - b =

A. 1/2^99

B. 1/9

C. 1/4

D. 3/2^99

E. 1/2^97
ShowHide Answer
Official Answer
D
User avatar
gmatophobia
User avatar
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 18 Nov 2025
Posts: 3,170
Own Kudos:
10,413
Given Kudos: 1,861
Location: India
Concentration: Strategy, Leadership
Posts: 3,170
Kudos: 10,413
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 21 Jan 2023, 12:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 bees. For every bee received, there is a 50% chance that it is a male and a 50% chance that it is a female. If a jar has at least 1 male, but more females than males, the honey farm keeps the jar; otherwise the jar is returned. If:

    a = the probability that Jar 1 is kept.b = the probability that Jar 2 is kept.

What is the value of a - b =

A. 1/2^99

B. 1/9

C. 1/4

D. 3/2^99

E. 1/2^97
Show more

Not sure what I am doing wrong, my answer doesn't match any of the options :(

Here is what I have done so far-

A jar can be accepted if it meets both the conditions

  1. At least one male bee
  2. Number of females > Number of males (Note that both the jars have odd number of bees, hence number of males cannot be equal to the number of females)

Probability (At least one male bee) = 1 - Probability(all female bees)

= \(1 - (\frac{1}{2})^{n}\)

n is the number of bees in the jar

Number of females> Number of males = I think the Probability of this is \(\frac{1}{2}\) (either the number of males is greater than the number of females or vice versa).

Jar 1

Probability of keeping the jar = a = Probability (At least one male bee) AND Probability of (Number of females> Number of males)

= \((1 - (\frac{1}{2})^{99}) * \frac{1}{2}\)

= \((\frac{1}{2} - (\frac{1}{2})^{100})\)

Jar 2

Probability of keeping the jar = b = Probability (At least one male bee) AND Probability of (Number of females> Number of males)

= \((1 - (\frac{1}{2})^{97} ) * \frac{1}{2}\)

= \((\frac{1}{2} - (\frac{1}{2})^{98})\)

a -b

\((\frac{1}{2} - (\frac{1}{2})^{100}) - ( \frac{1}{2} - (\frac{1}{2})^{98})\)

\((\frac{1}{2} - (\frac{1}{2})^{100}) - \frac{1}{2} + (\frac{1}{2})^{98}\)

\(\frac{1}{2}^{98} - \frac{1}{2}^{100}\)

\(\frac{1}{2}^{98}(1-\frac{1}{2}^{2})\)

\(\frac{1}{2}^{98}(1-\frac{1}{4})\)

\(\frac{1}{2}^{98}(\frac{3}{4})\)

\(\frac{3}{2^{100}}\)

Bunuel - Can you share some hint as to what I am doing incorrect here :problem:
User avatar
AzogTheFiler
User avatar
Joined: 15 Dec 2022
Last visit: 27 Jan 2023
Posts: 4
Own Kudos:
7
 [4]
Posts: 4
Kudos: 7
 [4]
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 22 Jan 2023, 00:00
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 bees. For every bee received, there is a 50% chance that it is a male and a 50% chance that it is a female. If a jar has at least 1 male, but more females than males, the honey farm keeps the jar; otherwise the jar is returned. If:

    a = the probability that Jar 1 is kept.

    b = the probability that Jar 2 is kept.

What is the value of a - b =

A. 1/2^99

B. 1/9

C. 1/4

D. 3/2^99

E. 1/2^97
Show more


P(Male) = P(Female) = \(\frac{1}{2}\)

Probability of keeping Jar 1
1/2^99 = a

Probability of keeping Jar 2
1/2^97 = b


a-b = 1/2^99 - 1/2^97

1/2^97(1/2^2 - 1)

1/2^97 (1/4 - 1)

1/2^97 (3/4)

1/2^97 (3/2^2)

3/2^99
User avatar
abhishekmayank
User avatar
Joined: 26 Apr 2016
Last visit: 28 Jan 2024
Posts: 201
Own Kudos:
59
 [1]
Given Kudos: 6
GMAT 1: 640 Q44 V33
GMAT 1: 640 Q44 V33
Posts: 201
Kudos: 59
 [1]
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 24 Jan 2023, 13:10
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Since in the jar a, at max 44 male bees could be for getting this jar selected then

p(a) = 1/2*1/2^98(probability of 1 male and 98 female bees) + 1/2^2*1/2^97(probability of 2 male and 97 female bees)
+ ......... + 1/2^44*1/2^45
= 44/2^99

Similarly, jar b can have minimum 1 and maximum 43 male bees, so

p(b) = 43/2^99

So probability of selecting jar a = p(a).(1-p(b)) (Jar a is selected and jar b is rejected)
and probability of selecting jar b = p(b).(1-p(a)) (Jar b is selected and jar a is rejected)

Where I am making mistake ?
User avatar
Regor60
User avatar
Joined: 21 Nov 2021
Last visit: 17 Nov 2025
Posts: 528
Own Kudos:
383
 [4]
Given Kudos: 459
Posts: 528
Kudos: 383
 [4]
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 25 Jan 2023, 15:57
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 bees. For every bee received, there is a 50% chance that it is a male and a 50% chance that it is a female. If a jar has at least 1 male, but more females than males, the honey farm keeps the jar; otherwise the jar is returned. If:

    a = the probability that Jar 1 is kept.

    b = the probability that Jar 2 is kept.

What is the value of a - b =

A. 1/2^99

B. 1/9

C. 1/4

D. 3/2^99

E. 1/2^97
Show more


The probability of a jar having more females than males is the same as the probability of having more males than females.

Since these are equal, each must be 1/2.

However, the probability of 1/2 for more females includes the disallowed case of 0 males, which has a probability of 1/2^99 for the first jar and 1/2^97 for the second.

So, A=1/2 - 1/2^99 and
B=1/2 - 1/2^97

So A-B = 1/2-1/2+1/2^97-1/2^99 =

(2^99-2^97)/2^99*2^97 =

2^97(2^2 - 1)/2^97*2^99 =

3/2^99

Posted from my mobile device
User avatar
Regor60
User avatar
Joined: 21 Nov 2021
Last visit: 17 Nov 2025
Posts: 528
Own Kudos:
383
 [2]
Given Kudos: 459
Posts: 528
Kudos: 383
 [2]
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 25 Jan 2023, 16:07
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AzogTheFiler



P(Male) = P(Female) = \(\frac{1}{2}\)

Probability of keeping Jar 1
1/2^99 = a

Probability of keeping Jar 2
1/2^97 = b


a-b = 1/2^99 - 1/2^97

Show more

This subtraction yields a negative number

Posted from my mobile device
User avatar
Regor60
User avatar
Joined: 21 Nov 2021
Last visit: 17 Nov 2025
Posts: 528
Own Kudos:
383
 [2]
Given Kudos: 459
Posts: 528
Kudos: 383
 [2]
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 27 Jan 2023, 08:51
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gmatophobia
Bunuel
Bunuel
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 bees. For every bee received, there is a 50% chance that it is a male and a 50% chance that it is a female. If a jar has at least 1 male, but more females than males, the honey farm keeps the jar; otherwise the jar is returned. If:

    a = the probability that Jar 1 is kept.

    b = the probability that Jar 2 is kept.

What is the value of a - b =

A. 1/2^99

B. 1/9

C. 1/4

D. 3/2^99

E. 1/2^97
________________________________________
Anyone want to try this ?

Not sure what I am doing wrong, my answer doesn't match any of the options :(

Here is what I have done so far-

A jar can be accepted if it meets both the conditions

  1. At least one male bee
  2. Number of females > Number of males (Note that both the jars have odd number of bees, hence number of males cannot be equal to the number of females)

Probability (At least one male bee) = 1 - Probability(all female bees)

= \(1 - (\frac{1}{2})^{n}\)

n is the number of bees in the jar

Number of females> Number of males = I think the Probability of this is \(\frac{1}{2}\) (either the number of males is greater than the number of females or vice versa).

Jar 1

Probability of keeping the jar = a = Probability (At least one male bee) AND Probability of (Number of females> Number of males)

= \((1 - (\frac{1}{2})^{99}) * \frac{1}{2}\)

= \((\frac{1}{2} - (\frac{1}{2})^{100})\)

Jar 2

Probability of keeping the jar = b = Probability (At least one male bee) AND Probability of (Number of females> Number of males)

= \((1 - (\frac{1}{2})^{97} ) * \frac{1}{2}\)

= \((\frac{1}{2} - (\frac{1}{2})^{98})\)

a -b

\((\frac{1}{2} - (\frac{1}{2})^{100}) - ( \frac{1}{2} - (\frac{1}{2})^{98})\)

\((\frac{1}{2} - (\frac{1}{2})^{100}) - \frac{1}{2} + (\frac{1}{2})^{98}\)

\(\frac{1}{2}^{98} - \frac{1}{2}^{100}\)

\(\frac{1}{2}^{98}(1-\frac{1}{2}^{2})\)

\(\frac{1}{2}^{98}(1-\frac{1}{4})\)

\(\frac{1}{2}^{98}(\frac{3}{4})\)

\(\frac{3}{2^{100}}\)

Bunuel - Can you share some hint as to what I am doing incorrect here :problem:
Show more


This is a conditional probability question.

Once it is assumed males>0, the probability of females>males changes from the initial probability as does the remaining space, both by (1/2)^99.

So it is the probability of females>males GIVEN males>0 that must be determined and then multiplied by the probability of >0 males.

This is

[(1/2)-(1/2)^99]/[1-(1/2)^99]

multiplied by 1-(1/2)^99.

The first simplifies to

[(2^98)-1]/[(2^99)-1]

and the second is equivalent to

[(2^99)-1]/(2^99)

So, when multiplied together yield

1/2 - (1/2^99)

Similarly for B

Posted from my mobile device
User avatar
gmatophobia
User avatar
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 18 Nov 2025
Posts: 3,170
Own Kudos:
10,413
 [1]
Given Kudos: 1,861
Location: India
Concentration: Strategy, Leadership
Posts: 3,170
Kudos: 10,413
 [1]
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 27 Jan 2023, 13:11
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Regor60

This is a conditional probability question.

Once it is assumed males>0, the probability of females>males changes from the initial probability as does the remaining space, both by (1/2)^99.

So it is the probability of females>males GIVEN males>0 that must be determined and then multiplied by the probability of >0 males.

This is

[(1/2)-(1/2)^99]/[1-(1/2)^99]

multiplied by 1-(1/2)^99.

The first simplifies to

[(2^98)-1]/[(2^99)-1]

and the second is equivalent to

[(2^99)-1]/(2^99)

So, when multiplied together yield

1/2 - (1/2^99)

Similarly for B

Posted from my mobile device
Show more

Thanks for responding! I liked your solution and it is way neater than mine.

Great approach :thumbsup:
User avatar
SKGM
User avatar
Joined: 27 Jun 2022
Last visit: 17 Sep 2024
Posts: 20
Own Kudos:
10
 [1]
Given Kudos: 34
Location: India
GMAT 1: 570 Q40 V28
GMAT 1: 570 Q40 V28
Posts: 20
Kudos: 10
 [1]
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 11 Jun 2023, 03:18
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avggmatstudent23
Bunuel can you please post the detailed solution? My answer is coming out to a negative number -3/2^99 (i.e. negative of option D). Not sure where I'm going wrong.
Show more

If you do not mind, my solution is following, using Bernoulli trials :

probality of at least one male in jar 1 & jar1 is kept, a: 1 - 99c0 (1/2)^0(1/2)^99
here, 99c0 (1/2)^0(1/2)^99 is probality with 0 male

= 1 - (1/2)^99
probality of at least one male in jar 2 & jar2 is kept, b: 1 - 97c0 (1/2)^0(1/2)^97
here, 97c0 (1/2)^0(1/2)^97 is probality with 0 male

= 1 - (1/2)^97

Now, value of a - b = [1 - (1/2)^99] - [1 - (1/2)^97]
= (1/2)^97 - (1/2)^99
= 3/2^99 .......option D :please:
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,588
Own Kudos:
1,079
Posts: 38,588
Kudos: 1,079
A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97 22 Nov 2024, 15:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Bunuel
Math Expert
105355 posts
Krunaal
Tuck School Moderator
805 posts