YOU HAVE 7 CANDELA LIT 2 OF THEM GO OUT HOW MA RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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Riddles and Answers © 2024

Found Beneath A Chimney

Hint:
A fireplace
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An Area With Little Rain

Hint:
They are deserts
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Two Camels Riddle

Hint:
The two camels were facing each other the entire time. Hence facing in opposite directions.
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Born In Mourning

Hint:
A tombstone
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Found In A Graveyard Riddle

Hint:
Tombstone
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Mortal Privation Riddle

Hint:
A tombstone
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Webbed Wings That Can Fly

Hint:
Its a bat
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Found In A Cave Riddle

Hint:
Bat
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The Youngest Level Of Girl Scouts

Hint:
A daisy
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A Character In Thomas And Friends

Hint:
Im a daisy
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The Walls Of Jericho

Hint:
Joshua
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Two Tablets Of Stone

Hint:
Moses
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The Loaded Revolver Riddle

Hint:
Henry should have Gretchen pull the trigger again without spinning.

We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.

If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
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Two In A Row Riddle

Hint: Who does he need to beat to win?
Father-mother-father

To beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second.
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The Secret Santa Exchange

Hint: It's not as difficult as it seems. It's the number of ways the friends can form a circle divided by the number of ways the names can be drawn out of the hat.
1/10

For a group of n friends, there are n! (n factorial) ways to draw the names out of the hat. Since a circle does not have a beginning and end, choose one person as the beginning and end of the circle. There are now (n-1)! ways to distribute the remaining people around the circle. Thus the probability of forming a single circle is

(n-1)! / n!

Since n! = (n-1)! * n (for n > 1), this can be rewritten as

(n-1)! / (n*(n-1)!)

Factoring out the (n-1)! from the numerator and denominator leaves

1/n

as the probability.
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