A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97

Question

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

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

Regor60 · Accepted Answer

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

gmatophobia · Answer

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 At least one male bee 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

AzogTheFiler · Answer

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

abhishekmayank · Answer

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 ?

Regor60 · Answer

This subtraction yields a negative number

Posted from my mobile device

Regor60 · Answer

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

/

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

The first simplifies to

/

and the second is equivalent to

/(2^99)

So, when multiplied together yield

1/2 - (1/2^99)

Similarly for B

Posted from my mobile device

gmatophobia · Answer

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

SKGM · Answer

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 = - = (1/2)^97 - (1/2)^99 = 3/2^99 .......option D

Kinshook · Answer

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

a = the probability that Jar 1 is kept.

b = the probability that Jar 2 is kept.

What is the value of a - b =

Jar 1: - 
 (1+1)^{99} = 99C0 + 99C1 + 99C2 + ... + 99C99 = 2*99C0 + .... + 2*99C49 
 99C0 + 99C1 + .... + 99C49 = \frac{2^{99}}{2} = 2^{98} 
 99C1 + .... + 99C49 = \frac{2^{99}}{2} -1 = 2^{98} - 1

Total ways =  2^99  
Favorable ways =  99C1 + 99C2 + ..... + 99C49 = \frac{2^{99}}{2} - 1 = 2^{98} - 1

a = the probability that Jar 1 is kept =  \frac{2^{98} - 1}{2^{99}} = \frac{1}{2} - \frac{1}{2^{99}}

Jar 2: - 
 (1+1)^{97} = 97C0 + ..... + 97C97 = 2*97C0 + .... 2*97C48 
 97C0 + 97C1 + .... + 97C48 = \frac{2^{97}}{2} = 2^{96} 
 97C1 + .... + 97C48 = 2^{96} - 1

Total ways =  2^97 
Favorable ways =  97C1 + ....+ 97C48 = 2^{96} - 1

b = the probability that Jar 2 is kept =  \frac{2^{96} - 1}{2^{97}} = \frac{1}{2} - \frac{1}{2^{97}}

a - b = (\frac{1}{2} - \frac{1}{2^{99}}) - (\frac{1}{2} - \frac{1}{2^{97}}) = \frac{1}{2^{97}} - \frac{1}{2^{99}} = \frac{2^2 - 1}{2^{99}} = \frac{3}{2^{99}}

IMO D

Hellohello007 · Answer

Hi,

How can we say that the probability of having more males than females in the jar is the same as having more females than males in the jar? Can someone please help?

Bunuel · Answer

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

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

Since each jar contains an odd number of bees and male/female are equally likely, the probability that females are in the majority is 1/2.

But the farm also requires at least 1 male, so we must exclude the case in which all bees are female.

For Jar 1:

a = 1/2 - 1/2^99

For Jar 2:

b = 1/2 - 1/2^97

Therefore:

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

= 4/2^99 - 1/2^99 =

= 3/2^99

Answer: D.

Because each bee is equally likely to be male or female. For every outcome with more females than males, there is an equally likely matching outcome with the genders reversed.

For example, 60 females and 39 males is just as likely as 60 males and 39 females.

Since the total number of bees is odd, a tie is impossible. So the two cases have equal probability, 1/2 each.

egmat · Answer

Hi Hellohello007,

You're looking at the line in Regor60's solution that says "the probability of more females than males is the same as more males than females, so each is 1/2." That's the step to unpack.

The key fact is that  each bee is 50% male and 50% female  - the two outcomes are perfectly interchangeable. Because of that, "male" and "female" are just labels with no bias between them. Whatever is true for one is equally true for the other.

Why the two probabilities must be equal

Take  any  outcome where there are more males than females. Now imagine flipping every bee's label - every male becomes a female and every female becomes a male. That new outcome now has  more females than males , and because each bee had an equal  1/2  chance either way, the flipped outcome is  exactly as likely  as the original.

So every "more males" outcome pairs up one-to-one with an equally likely "more females" outcome. The two piles have to weigh the same. Since the jars have an  odd  number of bees, a tie is impossible, so these two piles are the  only  possibilities and together they make  1 . Each must  therefore  be  1/2 .

Lock it in with a tiny jar

Shrink the problem to just  3 bees  and list all  8  equally likely outcomes (M/F for each):

- More males: MMM, MMF, MFM, FMM - that's  4 
- More females: FFF, FFM, FMF, MFF - that's  4

Exactly split,  4  and  4 , so each is  4/8  =  1/2 . The same balance holds for  99  or  97  bees - the count is huge, but the symmetry is identical.

That's why the solution can start from  1/2  for "more females," and then simply subtract the one banned case (the all-female jar) to get  a  and  b .

Answer: D

KarishmaB · Answer

Here is a post on symmetry that explains why the probability of more females than males is 1/2: https://anaprep.com/combinatorics-linea ... -symmetry/

A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97

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A honey farm orders 2 jars of bees. Jar 1 has 99 bees; jar 2 has 97

Prep Toolkit

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