A probability puzzle

[lorinda]

Ooops, once again I made an error, didnt read the bit that says I have 50 coins....

Perhaps a holiday is in order [img]/forums/images/icons/frown.gif[/img]

[Bozeman]

Let me try.

Suppose you have x dimes, y nickels, z pennies.
You can't have 4 quarters, or you'd only have 4 coins.

If you have 3 quarters, x+y+z=49, 10x+5y+z=25. Since 25<49, this has no all non-negative solutions.

If you have two quarters, x+y+z=48, 10x+5y+z=50. Thus 9x+4y=2, impossible for any integers x,y.

If you have one quarter, x+y+z=49, 10x+5y+z=75. 9x+4y=26. x=2, y=2 and then z=45 is the only integer solution.

0 quarters, x+y+z=50, 10x+5y+z=100. 9x+4y=50.
x=2, y=8, z=40 is the only one that works.

If the probability of dropping is independent of type of coin, and each combination is equally likely, then P=0.5*1/50=1%

For me, chance of dropping is inversely porportional to value, so P=0.5*c*1/25 where 1=c*sum(1/value), so c=1/45.64, and P~.044% [img]/forums/images/icons/smile.gif[/img]

Craig

[JTG51]

Of course, the answer many of you came up with is the correct one. There is a 1/100 chance of losing the quarter.

[Ray Zee]

i assume too much. when i first did it i thought it natural that the question did assume you had a quarter in the 50 coins, so i had to look twice when i saw 1 in 100 for an answer. but then i realized that for some people the answer had to be zero. know why