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:

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

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

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

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

bumpbot · Answer

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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

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