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Why the probability of picking any ball numbered (1 to 4) in ONE draw

out of a box containing 37 balls which is (4/37)= (0.108108), DOES NOT equal to the probability of picking any ball numbered (1-2) at LEAST once[with replacement] out of the same box of 37 balls in TWO draws which is [1-(35/37)x(35/37)]=( 0.106975)???

I mean In the first case we are talking about picking a ball from 4 balls. In the second case, even though we HALVED the number of balls to choose out of from (4 to 2) we DOUBLED the number of draws, should it that make both probabilities EQUAL??? Any MATHEMATICAL explaination!!!!!!!

out of a box containing 37 balls which is (4/37)= (0.108108), DOES NOT equal to the probability of picking any ball numbered (1-2) at LEAST once[with replacement] out of the same box of 37 balls in TWO draws which is [1-(35/37)x(35/37)]=( 0.106975)???

I mean In the first case we are talking about picking a ball from 4 balls. In the second case, even though we HALVED the number of balls to choose out of from (4 to 2) we DOUBLED the number of draws, should it that make both probabilities EQUAL??? Any MATHEMATICAL explaination!!!!!!!

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P(A) + P(B) - P(A and B) = 2/37 + 2/37 -(2/37)^2 = .105186

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Start your 7-day free trial(4/37)=.108108108108...

but to get it equal you would have to take not only half as many balls your looking for but also half as many balls in the bag or whatever it is...so it would be like this:

(2/(37/2))=.108108108108..

--------------------------

here is an example to prove this works:

(11/22)=.5

now to make it equal the next equation it would be:

((11/2)/(22/2))= .5

so it works for any numbers and this could also be proved algebraically

--------------------------

(a/b)=c

((a/2)/(b/2))=c

therefore: (a/b) has to equal ((a/2)/(b/2))

using the simple formula ((a/b)/(c/d))=(ac/bd)

you can conclude that ((a/2)/(b/2)) = (2a/2b) which equals (a/b)... PROVEN

--------------------------

so that is how it is supposed to be done to make them equal each other...i think the formula is like "Product of the means over the product of the extremes"... well that is how i was always told to remeber it... everything comes back to it in this problem

hope this helps...post any more questions here

Mark

Mayank.

Now for Mr.simpsons17371, Of course 12/24=6/12=3/6. BUT the question is why is it when we HALVE the number of balls and DOUBLE the number of draws, we do not get EQUAL probabilites of the original experiment?? I am looking for the mathematical relation between Number of balls AND number of DRAWS!!!!!!!

33/37 chances of drawing 0 balls + 4/37 chances of drawing 1 ball = .108108

vs

1225/1369 chances of drawing 0 balls + 140/1369 chances of drawing 1 ball + 4/1369 chances of drawing 2 balls = .108108

The difference is that, when you try to pick one of 4 balls, you're only picking once. When you are twice picking 1 of 2 balls you can double up on the possible draws. These doubled up ones have to be discarded, since this CANNOT happen if you only pick once.

Note that 2 * (1 - 35/37) = 0.108108, i.e. the same as 1 - 33/37

The mathematical relationship is as I have stated: the probability of picking (1-2) in the first draw, PLUS the probability of picking (1-2) in the second draw, MINUS the probability of doubling up on the picks.

If you pick twice, with replacement. then the chance of getting any ball 1-19 cannot be 1/2 + 1/2 = 1

it is 1/2 + 1/2 - 1/2 * 1/2 = 3/4

To calculate the chances of picking balls 1-4 out of 37 balls

(4/37)=.1081081

or 1-(33/37)=.1081081

To calculate the chances of picking balls 1-2 out of 37 AT LEAST ONCE with replacement:

(1-(35/37)^2)=.105186

If the formula were (1-(35/37))*2 then the probability would be equal to 1-(33/37) and equal .1081081, but that isn't the mathmatical way of calculating the probability of multiple events.

I think the best way to look at the issue is to see if there is any mathematical explaination, or otherwise, for the fact that probabilities DO NOT function in a linear and proportional manner. Is there any explaination for that FACT!!!

e.g balls numbered 1-37 - draw one ball, chance of it being 1 = 1/37, draw two balls at once, prob = 2/37

up until if you draw all 37 balls at once, the probability of getting ball 1 = 37/37 = 1

now if you draw one ball at a time 37 times, the probability cannot be 1 - because how can you guarantee that 1 would have been drawn.

in fact it is 1 - (36/37)^37 = .637

I think what you are looking for is an intuitive reason, not a mathematical reason - I think what you need to accept there is that drawing two balls at once is a different scenario to making two separate drawings, because you are replacing the ball drawn in the separate drawings case

when drawing two balls at once you are eliminating the first ball drawn from the second draw - increasing your chance of success, hence the probability is higher.

balls numbered 3 and 4 stand out and

are more likely to be drawn than balls

numbered 1 and 2. hence the higher

probability for the single draw of balls

(1-4) case.

disproving this statement is left as an

excercise for the student.

leo

I did take another look at the question

and was going to do a simulation to get

results.... but in thinking about the design

of a simulation it became obvious that you should

not expect the same probability for both cases.

in case one you draw one number...in case two you draw

2 numbers ... in case 2 you could get the same number

twice.

work out the probability of getting the same number

twice for case two and look at that probability with

respect to your case 1, case 2 probabilities.

leo

you are correct in the first calculation:

the posibility of picking a ball numbered 1 to 4 from 37 balls

= 4/37

= 0.108108108

Very sorry to say, your second calcualtion( [1-(35/37)x(35/37)]=0.1069

Posibility of picking ball numbered 1 to 2 in both picks with replacement

= (1 - (35/37)) * (1 - (35/37))

= (2/37) * (2/37)

= 0.00292184

Posibility of picking ball numbered 1 to 2 in at least 1 pick with replacement

= (2/37) + (2/37)

= 2 * (2/37)

= 4/37

= 0.108108108

Conclusion:

the Posibility of picking ball numbered 1 to 2 in at least 1 pick in 2 pick with replacement IS SAME with posibility of picking ball numbered 1 to 4 in 1 pick.

>>the Posibility of picking ball numbered 1 to 2 in at least 1 pick in 2 pick with replacement IS SAME with posibility of picking ball numbered 1 to 4 in 1 pick

no its not

first scenario = 1 - (35/37)^2 =.10519

2nd scenario = 4/37 = .108108

To deighton, first scenario is not true, let me try to prove it:

posibility of pick up ball numbered 1 to 2 in 1 pick

= 2/37

since it is with replacement, the second pick should have same posiblity of picking up the ball numbered 1 to 2

= 2/37

both pick have posibilty of 2/37, so we can conclude that:

Posibility of picking ball numbered 1 to 2 in both picks with replacement

= (2/37) * (2/37)

= 0.00292184

Posibility of picking ball numbered 1 to 2 in at least 1 pick out of 2 picks with replacement

= 2 * (2/37)

= 4/37

= 0.108108108

<> 1 - (35/37)^2

Pls prove that I am wrong and not just telling me I am wrong. Thank You.

The following equalation is wrong since I miss out a fact that once you get the ball in first pick, you wont go for the second. perhaps this is the reason of the different of posibility in 2 different scenerios.

//THIS IS WRONG, PLS IGNORE

Posibility of picking ball numbered 1 to 2 in at least 1 pick out of 2 picks with replacement

= 2 * (2/37)

= 4/37

= 0.108108108

<> 1 - (35/37)^2

The correct calculation should be like this:

posibility of get balls number 1 to 2 inat least 1 pick out of 2 picks

= posibility of get in 1st pick + (not get in 1st pick * get in 2nd pick)

= 2/37 + (35/37 * 2/37)

= 0.105186267

Since the posibility of go for 2nd pick in 2 picks scenario is less than 1.0, and the posibility of go for 1st pick in 1 picks scenarion is 1.0. So the possibility of 2 picks scenarion is little bit lower.

Hope this time i am correct.Thanks.

1. If you calculated it on paper and not strait away calculate with calculator, you will find the following result:

1 - (35/37)^2 = 144/1369

2/37 + (35/37 * 2/37) = 144/1369

both of tem is equal to 0.105186267

2. What I means by less than 1.0 and equal to 1.0 is that:

In the 1 pick scenarion, it is 100% you will utilise all your chances to pick up a ball (1 pick in the 1 pick scenario)

In the 2 pick scenarion, it is NOT 100% you will utilise all your chances to pick up a ball (2 picks in the 2 pick scenario). The posibility is less than 100% since you might get the ball in first draw and stop without 2nd attempt.

So the posibility of 2 picksscenario is lower.

Hope this make it clear. Thanks.

= 1.0 -

Instead of 37 balls, let's use just 4. In that case, the probability of picking balls (1-4) in 1 draw is obviously 1. Equally obviously, there us no guarantee that in 2 draws we will pick balls (1-2) at least once (might might pick balls 3 or 4 both times). Hence that probability HAS to be less than 1.

Make sense?

Posibiblity of AT LEAST ONE

= get in 1st draw only + get in 2nd draw only + get in both draws

= (2/37 * 35/37) + (35/37 * 2/37) + (2/37 * 2/37)

= (70 + 70 + 4)/1369

= 144/1369

= 0.105186267

...Bill

I hope this calculation bring inspriration to others expert to solve the problem.

1 pick scenario

= 1 - 33/37

= 4/37

2 picks scenario

= 1 - (35/37)^2

= 1 - 1225/1369

= 144/1369

Since the prosibility of not get the balls is different

33/37 <> 1225/1369

So the posibility on get the balls is also different.

But why? Why the prosibility of not get the balls is different?

In the first case you have 37 possible outcomes, of which only 4 are acceptable. In the second, with *two* drawings from the same set, you have 37^2 (1369) possible outcomes, of which 144 are acceptable.

The possible outcomes of the second scenario are, if we designate Y as drawing 1 or 2, and N as drawing a ball numbered between 3 and 37

N&N: 1225 = 35X35

N&Y: 70 = 35X2

Y&N: 70 = 2X35

Y&Y: 4 = 2X2

Another way to look at it is to say that, in scenario #1, the probability of *not* succeeding is 33/37, and in scenario #2 it is (35^2)/(37^2). The probability of not succeeding in two drawings is the product of the probabilities of not succeeding in one drawing, not the sum.

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