BBB

BBG

BGG

 

Keep going... There are some other sample points(consider order).

Good guess. But no.

Good guess. But no.

 

Which guess?

Keep going... There are some other sample points(consider order).

 

Why would order matter? I know there's 3. I know one is a boy. What are the other possibilities?

Why would order matter? I know there's 3. I know one is a boy. What are the other possibilities?

 

If we didn't have any pre-knowledge of at least one boy and just knew he had three kids, the sample space would be:

 

S = { BBB, BBG , BGB, GBB , GBG, BGG, GGB, and GGG}

 

P( all boys) would be 1/8.

 

However, since we know he has at least one boy, the GGG sample point is eliminated from the sample space. Hence, the sample space for this conditional probably would include only seven sample points.

P(three boys) = P(BBB) = 1/7

If we didn't have any pre-knowledge of at least one boy and just knew he had three kids, the sample space would be:

 

S = { BBB, BBG , BGB, GBB , GBG, BGG, GGB, and GGG}

 

P( all boys) would be 1/8.

 

However, since we know he has at least one boy, the GGG sample point is eliminated from the sample space. Hence, the sample space for this conditional probably would include only seven sample points.

P(three boys) = P(BBB) = 1/7

 

Defies logic.

 

He can either have

 

3 boys

2 boys and 1 girl

1 boy and 2 girls

Hint: Let BL1 = blue on first draw and BL2 = blue on second draw. P(both are blue) = P(BL 1 and B2)= P(BL 1) * P( BL2 given BL1) = 2/3. So how many blue socks must there be so that P(BL1 and BL2) = 2/3 ?

 

For instance if there were 4 blue socks, P(B1 and B2) =4/6 * 3/5 = 2/5 and

P( BR 1 and BR 2) = 2/6 * 1/5 = 1/15 . So it can't be four blue and 2 brown since 2/5 is not = 2/3 . Try another possibility.

Defies logic.

 

He can either have

 

3 boys

2 boys and 1 girl

1 boy and 2 girls

 

 

So if I posed the question " A couple has three children, what is the probability that two of the kids are boys? , " what would the answer be?

I'll leave you all with this one. Made this one up a few years ago.

Mismatched Joe is in a pitch dark room picking out socks from a dresser drawer. There is a total of six socks in the drawer. There are only blue and brown socks. He can't tell anything about the color by just the feel of the sock. He chooses two socks at random. If he has a 2/3 chance of picking out a pair of blue socks, what is the chance he chooses a pair of brown socks?

 

If he is wearing long pants it won't matter, no one will know the difference. :lol2:

Hint: Let BL1 = blue on first draw and BL2 = blue on second draw. P(both are blue) = P(BL 1 and B2)= P(BL 1) * P( BL2 given BL1) = 2/3. So how many blue socks must there be so that P(BL1 and BL2) = 2/3 ?

 

For instance if there were 4 blue socks, P(B1 and B2) =4/6 * 3/5 = 2/5 and

P( BR 1 and BR 2) = 2/6 * 1/5 = 1/15 . So it can't be four blue and 2 brown since 2/5 is not = 2/3 . Try another possibility.

 

I need a drink.

For three children(with no pre-knowledge) the probabilities are as follows:

P(all boys) = 1/8

P(exactly two boys) = 3/8

P(at least two boys) = 4/8=1/2

P(at least one boy) = 7/8

P( no boys) = 1/8

These all follow from the fact that the sample space S={BBB, BBG , BGB, GBB, BGG, GBG, GGB,GGB) and each of the eight sample points are equally likely at 1/8. Hence, P(exactly two boys) = 1/8 + 1/8 + 1/8 = 3/8 .

 

Now, if we know that at least one of the kids is a boy, then when figuring the probability the GGG sample point does not figure into the sample space. We would have seven equally likely possibilities. Hence, P(all boys) = 1/7 . A good puzzle is one that seems to defy logic and is I counter-intuitive. This one, I think, is an example. Like I said, I have had people argue up and down about this one but the conditional probability upon which it is based is mathematically sound.

I need a drink.

 

You 13 and you imbibe devilish liquids? :stop:

Imagine that the distance from the earth to the sun (93 million miles, or about 8 light minutes) is compressed to the thickness of a typical sheet of paper. On this scale, the nearest star (Proxima Centauri @4.3 light years) is at a distance of 71 feet. The diameter of the Milky Way (100,000 light years) would require a 310 mile high stack of paper, while the distance to the Andromeda galaxy (at 2 million light years , one of the most distant objects visible to the naked eye) would require a stack of paper more than 6000 miles high! On this scale, the "edge" of the Universe, defined as the most distance known galaxy ever discovered(EGSY2008532660), some 13.25 billion light years away, is not reached until the stack of paper is almost 41.5 million miles high--nearly half of the way to the sun on the real scale of things! WOW! I LOVE SCIENCE!

The universe is so vast that it's mind boggling.

The universe is so vast that it's mind boggling.

 

Isn't it though? I'm thankful to be just a minute part of it!

One of the greatest mathematicians ever, a man by the name of Carl Friedrich Gauss, discovered a formula at an early age that allowed him to find the sum of the first n positive integers. For example , as a class exercise, when asked to find the sum of the first 100 positive integers(probably a "busy work" assignment from the teacher), young Gauss thought of it this way:

 

1+ 2+ 3+ 4+ 5+ 6+ 7 + ... + 97 +98+99 + 100

 

He noticed that if you paired up the numbers as follows, you always got the same sum:

1+100 = 101 ; 2+99=101 ; 3+98=101; etc... Therefore, there would be 50 such pairs of sums. Hence, 50* 101 = 5,050.

By this method, if you were adding the first 300 positive integers, you would have 150 pairs of numbers with each pair's sum being 301. Thus, the sum of the first 300 positive integers is 150*301 = 45,150

If you had an odd number of integers, say 101, then you would have 50 pairs( 101+1, 100+2, 99+3, ... ) each with a sum of 102 plus one lone number in the middle that had nothing to pair up with. That middle number is 102/2 or 51. Thus, the sum would be (50* 102) +51 = 5,151.

Little Carl then deduced a general formula for the sum of the first n positive integers. For an even number of integers : n/2 pairs each having a sum of (n+1). Therefore Sum = n(n+1)/2 . If there is an odd number of integers: (n-1)/2 pairs ,each having a sum of (n+1) PLUS that middle number that wasn't paired up, which would be (n+1)/2 . Thus , S = [ (n-1)/2] *[(n+1) ] + [(n+1)/2] =

[(n^2 -1) /2] + [(n+1)/2] = (n^2 -1 +n+1)/2 = (n^2 +n)/2 = n(n+1)/2 . SAME ANSWER! So, whether you are adding an even number of positive integers or an odd number of positive integers, the sum is the same.... n(n+1)/2 !

Oh .... by the way, young Carl was in early elementary school when he discovered this. I mean, like, second or third grade!!!

 

Can you imagine what a little bastard :watching: his little schoolmates thought he was ? Took them all class period and took him maybe twenty seconds to get the answer.

Edited November 25, 201510 yr by Science Friction