Problem with solving a Probability problem

fendi

fendi

fendi

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23 May 2005
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2,156
Hi dudes,

I need to work out a probability outcome of a scenario:

I have 300 balls, 290 different and 10 red. I have two chance of picking the red balls, with an extra bonus chance if I can't find two. so it will be either 2 chances if I get both right, or 3 if I only get 1 right in the first 2.

Please please help as my math sucks.

Thanks a lot every one.
 
Assuming 0.9r=1, then you'll find that the aeroplane on the treadmill will remain stationary, the goat will eat grass in in the radius of the pole to which it is tied and even after opening the first door, you still have a 1/3 chance of picking the Cadillac from the remaining two doors.

Then all you need to do is take that answer and divide it by 0. Bob's your uncle, Fanny's your aunt.

Also, based on how I read what you wrote, your answer seems correct to me.
 
Theophany said:
Assuming 0.9r=1, then you'll find that the aeroplane on the treadmill will remain stationary, the goat will eat grass in in the radius of the pole to which it is tied and even after opening the first door, you still have a 1/3 chance of picking the Cadillac from the remaining two doors.

Then all you need to do is take that answer and divide it by 0. Bob's your uncle, Fanny's your aunt.

Also, based on how I read what you wrote, your answer seems correct to me.
Click to expand...

Does this take into account variable change if I switch from first door to the second?
 
Theophany said:
Assuming 0.9r=1, then you'll find that the aeroplane on the treadmill will remain stationary, the goat will eat grass in in the radius of the pole to which it is tied and even after opening the first door, you still have a 1/3 chance of picking the Cadillac from the remaining two doors.

Then all you need to do is take that answer and divide it by 0. Bob's your uncle, Fanny's your aunt.

Also, based on how I read what you wrote, your answer seems correct to me.
Click to expand...

And yet you'll still only have 1 sandwich, no matter how many times you cut it in half.
 
So the first ball you pick, there is a 10/300 chance of getting a red ball = 0.03333

If you get a red ball, then on the second there is a 9/299 chance of getting a red ball = 0.03010

If you do not get a red ball first time round, then second time there is a 10/299 chance = 0.03344

If you get one red from your first two, the chance of getting a red in the third is 9/298 = 0.03020

So, the probability of you getting two reds in the first two draws is: 0.03333x0.03010=0.001003233, or 1%

The probability of you getting a red on the first and third draw is: 0.03333x0.03020=0.001006566 (or 1% again!) - very, very marginally higher

The probability of you getting a red on the second and third draw is: 0.03344x0.03020=0.0010098888, which is again very slightly higher.

I *think* that's right, but you'll want to get someone who's better at this than me to check it!
 
manic111 said:
So the first ball you pick, there is a 10/300 chance of getting a red ball = 0.03333

If you get a red ball, then on the second there is a 9/299 chance of getting a red ball = 0.03010

If you do not get a red ball first time round, then second time there is a 10/299 chance = 0.03344

If you get one red from your first two, the chance of getting a red in the third is 9/298 = 0.03020

So, the probability of you getting two reds in the first two draws is: 0.03333x0.03010=0.001003233, or 1%

The probability of you getting a red on the first and third draw is: 0.03333x0.03020=0.001006566 (or 1% again!) - very, very marginally higher

The probability of you getting a red on the second and third draw is: 0.03344x0.03020=0.0010098888, which is again very slightly higher.

I *think* that's right, but you'll want to get someone who's better at this than me to check it!
Click to expand...

You got all the hard bits right. 0.001003233 is around 0.1%, not 1%. The same applies for the others. :)
 
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