These are the whimsical thoughts that I shall be having during my times at university...
Wednesday, 23 February 2011
Give this one a try...
“A vagrant who stole three melons weighing three pounds each, came to a bridge which was just strong enough to hold him and six pounds. Without throwing any of the melons across the bridge, how did the vagrant cross the bridge with the melons, none of which touched the bridge?”
Sunday, 20 February 2011
The Answer!!!!!
How can you put 10 lumps of sugar into three cups so there is an odd number of lumps in each cup?
Put five lumps in the first cup, two in the second, and three in the third.
Then sit the third cup in the second cup.
:D :D :D :D :D
Thursday, 17 February 2011
More maths!!!
Its been a while since I've posted on here - the dissertation is really dragging me down, plus I went away to snowdon for a weekend's climbing :D
Heres a quick one today, promise ill put a better one up next time :D
How can you put 10 lumps of sugar into three cups so there is an odd number of lumps in each cup?
Heres a quick one today, promise ill put a better one up next time :D
How can you put 10 lumps of sugar into three cups so there is an odd number of lumps in each cup?
Tuesday, 1 February 2011
The Answer!!!
I'm sure a lot of you are going to be annoyed by this.... but it is just a simple case of where common sense is wrong :)
I have two children, and at least one of them is a son. What is the probability that the other is also a son?
If I have only one child, the probability is $\frac{1}{2}$ that it’s a boy.
But if I have two children, there are really four possibilities, depending on birth order: boy-boy, boy-girl, girl-boy, girl-girl. You know that at least one child is a boy, so you can exclude the girl-girl option. Of the three that remain, only one has two boys. So the probability that my other child is a boy is $\frac{1}{3}$.
Ta-dah!
I have two children, and at least one of them is a son. What is the probability that the other is also a son?
If I have only one child, the probability is $\frac{1}{2}$ that it’s a boy.
But if I have two children, there are really four possibilities, depending on birth order: boy-boy, boy-girl, girl-boy, girl-girl. You know that at least one child is a boy, so you can exclude the girl-girl option. Of the three that remain, only one has two boys. So the probability that my other child is a boy is $\frac{1}{3}$.
Ta-dah!
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