Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide?
So let’s think this through. The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anti-clockwise). If the ants do not pick the same direction, there will definitely be a collision. Each ant has the option to either move clockwise or anti-clockwise. There is a one in two chance that an ant decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision.
P(No collision) = P(All ants go in a clockwise direction) + P( All ants go in an anti-clockwise direction) = 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25cleanfunny
A bus driver was heading down a street in Colorado. He went right past a stop sign without stopping, he turned left where there was a "no left turn" sign, and he went the wrong way on a one-way street. Then he went on the left side of the road past a cop car. Still - he didn't break any traffic laws.
He was walking...not driving.logic
A deliveryman comes to a house to drop off a package. He asks the woman who lives there how many children she has.
"Three," she says. "And I bet you can't guess their ages."
"Ok, give me a hint," the deliveryman says.
"Well, if you multiply their ages together, you get 36," she says. "And if you add their ages together, the sum is equal to our house number."
The deliveryman looks at the house number nailed to the front of her house. "I need another hint," he says.
The woman thinks for a moment. "My youngest son will have a lot to learn from his older brothers," she says.
The deliveryman's eyes light up and he tells her the ages of her three children. What are their ages?
Their ages are 1, 6, and 6. We can figure this out as follows:
Given that their ages multiply out to 36, the possible ages for the children are:
1, 1, 36 (sum = 38)
1, 2, 18 (sum = 21)
1, 3, 12 (sum = 16)
1, 4, 9 (sum = 14)
1, 6, 6 (sum = 13)
2, 2, 9 (sum = 13)
2, 3, 6 (sum = 11)
3, 3, 4 (sum = 10)
When the woman tells the deliveryman that the children's ages add up to her street number, he still doesn't know their ages. The only way this could happen is that there is more than one possible way for the children's ages to add up to the number on the house (or else he would have known their ages when he looked at the house number). Looking back at the possible values for the children's ages, you can see that there is only one situation in which there are multiple possible values for the children's ages that add up to the same sum, and that is if their ages are either 1, 6, and 6 (sums up to 13), or 2, 2, and 9 (also sums up to 13). So these are now the only possible values for their ages.
When the woman then tells him that her youngest son has two older brothers (who we can tell are clearly a number of years older), the only possible situation is that their ages are 1, 6, and 6.logicmathshort
Take 9 from 6, 10 from 9, 50 from 40 and leave 6. How is it possible?
SIX - 9 (IX) = S
9 (IX) - 10 (X) = I
40 (XL) - 50 (L) = X