Challenge your brains! Can you solve this problem?

[Fan from VT]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

50%?

[Bahku]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

[Roy Hobbs]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

Can you show your work? 

My thought was the same as Fan from VT's; the gender of the second dog has nothing to do with the gender of the male dog (i.e., it's an independent event).  Thus, it can be male or female, so it's 50/50, right?

[GKC]

The ball problem reminded me of something the Dallas Mavericks stats guy said. You always shoot a 3 to win especially when away, when down 2. Reason being, is that you have say 35% to hit a 3. 50% to hit a 2, but usually 50% chance to win overtime (home teams win slightly more). So if you take a 2, you only have about 25% chance to win the game.

Obviously foul trouble and momentum should be looked at, but all things equal, go for the win.

heres one...

One machine can shred a truck load of paper in 4 hours, and another machine can shred a truck load of paper in 2 hours. If the machines were to shred together, how long would it take to shred a truck load of paper?

Truck A = X
Truck B = Y
Z = Minutes
A = How many truckloads of paper is shredded
X shreds 1/240 truck loads of paper per minute (240 minutes in 4 hours)
Y shreds 1/120 truck loads of paper per minute (120 minutes in 2 hours)

So A = (1/240)z + (1/120)z
If Z = 1, A = 0.0125
If A = 1, Z = 80

So 80 minutes.

---

This one's one I found in class:

You have 2 pieces of string.
Each, when lit, will burn from one end to another in an hour.
However, the interval in which it burns is random (ie it could burn faster at certain points, slower at others. Though total it will take an hour).

How do you use the 2 pieces of string and a lighter to measure 45 minutes?

Light one string at one end and at the same time light the other string at both ends.

When the string that is doubly lit completely burns out (i.e. the flames meet), it's been 30 minutes. At that moment, light the unlit end of the other string, the one burning at only one end. When those two flames meet and extinguish the second string, it's been 45 minutes

Excellent job. That's a tough one. TP

[Bahku]

Here's one: The professor gave one of the students a 40 pound brick and told him to cut it up into 4 separate pieces so that it would be possible to weigh any amount up to the 40 pounds. The student came back to class only to tell the professor that he couldn't figure out how to do it and then accidentally dropped it on the floor and low and behold it broke into the 4 exact pieces. What was the weight of each piece?

1 lb, 3 lbs, 9 lbs, and 27 lbs. use some multiples to add their weight, or put some on the other side of the scale to subtract them.

[GKC]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

Can you show your work? 

My thought was the same as Fan from VT's; the gender of the second dog has nothing to do with the gender of the male dog (i.e., it's an independent event).  Thus, it can be male or female, so it's 50/50, right?

Exactly what I thought. It's an independent event so it should be 50%

[Master Po]

Why bother to solve this......just Ask Edgar in that thread.....my brain needs the rest

[Bahku]

heres one...

One machine can shred a truck load of paper in 4 hours, and another machine can shred a truck load of paper in 2 hours. If the machines were to shred together, how long would it take to shred a truck load of paper?

Rondo is correct on this one ... 1 hour and 20 minutes.

[Redz]

Here's one: The professor gave one of the students a 40 pound brick and told him to cut it up into 4 separate pieces so that it would be possible to weigh any amount up to the 40 pounds. The student came back to class only to tell the professor that he couldn't figure out how to do it and then accidentally dropped it on the floor and low and behold it broke into the 4 exact pieces. What was the weight of each piece?

1 lb, 3 lbs, 9 lbs, and 27 lbs. use some multiples to add their weight, or put some on the other side of the scale to subtract them.

Nice one.  Though I suppose the wording of the question should be "any EVEN amount".

[Fan from VT]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

Can you show your work? 

My thought was the same as Fan from VT's; the gender of the second dog has nothing to do with the gender of the male dog (i.e., it's an independent event).  Thus, it can be male or female, so it's 50/50, right?

Exactly what I thought. It's an independent event so it should be 50%

I think it's right. it's like the question:
I'm flipping a coin. I just got heads. What are the chances the next flip will be heads?

or

I'm flipping a coin. I've gotten heads 999 times in a row. what are the odds the next flip will be heads?


Answer in both: 50%

[Bahku]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

Can you show your work? 

My thought was the same as Fan from VT's; the gender of the second dog has nothing to do with the gender of the male dog (i.e., it's an independent event).  Thus, it can be male or female, so it's 50/50, right?

Exactly what I thought. It's an independent event so it should be 50%

Say you were picking them randomly, these would be the variables:

Female/Female
Female/Male
Male/Female
Male/Male ...

with the info gievn you are eliminating the last: Male/Male (because the probability being decided, it's 100% Male) ...

 ... you are then left with:

Female/Female
Female/Male
Male/Female

.. this leaves a 1 in 3 chance of the determining the other sex.

[Roy Hobbs]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

Can you show your work? 

My thought was the same as Fan from VT's; the gender of the second dog has nothing to do with the gender of the male dog (i.e., it's an independent event).  Thus, it can be male or female, so it's 50/50, right?

Exactly what I thought. It's an independent event so it should be 50%

Say you were picking them randomly, these would be the variables:

Female/Female
Female/Male
Male/Female
Male/Male ...

with the info gievn you are eliminating the last: Male/Male (because the probability being decided, it's 100% Male) ...

 ... you are then left with:

Female/Female
Female/Male
Male/Female

.. this leaves a 1 in 3 chance of the determining the other sex.

The question says "what are the odds that the other one is male?"  I.e., what are the chances that the dog you don't know the gender of is male?

In other words, Dog #1 = male.  Thus, you only have two other outcomes:  male or female.  Thus, it's 50%.

[bdm860]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

Can you show your work? 

My thought was the same as Fan from VT's; the gender of the second dog has nothing to do with the gender of the male dog (i.e., it's an independent event).  Thus, it can be male or female, so it's 50/50, right?

Exactly what I thought. It's an independent event so it should be 50%

Say you were picking them randomly, these would be the variables:

Female/Female
Female/Male
Male/Female
Male/Male ...

with the info gievn you are eliminating the last: Male/Male (because the probability being decided, it's 100% Male) ...

 ... you are then left with:

Female/Female
Female/Male
Male/Female

.. this leaves a 1 in 3 chance of the determining the other sex.

Why do you eliminate the Male/Male option? 

[Bahku]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

Can you show your work? 

My thought was the same as Fan from VT's; the gender of the second dog has nothing to do with the gender of the male dog (i.e., it's an independent event).  Thus, it can be male or female, so it's 50/50, right?

Exactly what I thought. It's an independent event so it should be 50%

Say you were picking them randomly, these would be the variables:

Female/Female
Female/Male
Male/Female
Male/Male ...

with the info gievn you are eliminating the last: Male/Male (because the probability being decided, it's 100% Male) ...

 ... you are then left with:

Female/Female
Female/Male
Male/Female

.. this leaves a 1 in 3 chance of the determining the other sex.

The question says "what are the odds that the other one is male?"  I.e., what are the chances that the dog you don't know the gender of is male?

In other words, Dog #1 = male.  Thus, you only have two other outcomes:  male or female.  Thus, it's 50%.

No ... only if you're considering the sex of one ... this male/female paradox has been studied and investigated extensively, and the answer depends completely on how the information is given. In this case, the info is presented with two variables: Male and Male, thus eliminating the first option of 4 variants. It leaves three ... and the answer is 1/3.

[Roy Hobbs]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Algebraically, the answer has to be 1 in 3.

Can you show your work? 

My thought was the same as Fan from VT's; the gender of the second dog has nothing to do with the gender of the male dog (i.e., it's an independent event).  Thus, it can be male or female, so it's 50/50, right?

Exactly what I thought. It's an independent event so it should be 50%

Say you were picking them randomly, these would be the variables:

Female/Female
Female/Male
Male/Female
Male/Male ...

with the info gievn you are eliminating the last: Male/Male (because the probability being decided, it's 100% Male) ...

 ... you are then left with:

Female/Female
Female/Male
Male/Female

.. this leaves a 1 in 3 chance of the determining the other sex.

The question says "what are the odds that the other one is male?"  I.e., what are the chances that the dog you don't know the gender of is male?

In other words, Dog #1 = male.  Thus, you only have two other outcomes:  male or female.  Thus, it's 50%.

No ... only if you're considering the sex of one ... this male/female paradox has been studied and investigated extensively, and the answer depends completely on how the information is given. In this case, the info is presented with two variables: Male and Male, thus eliminating the first option of 4 variants. It leaves three ... and the answer is 1/3.

We are only considering the sex of one:  "the other one".  We don't know which dog is the one that is definitely male, but we don't have to.  We just need to know the gender of the "other" one.  The "other" one can only be one of two genders.