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Pennies and Dimes

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by logitech » Tue Nov 25, 2008 10:44 pm
A hand purse contains 6 nickels, 5 pennies and 4 dimes. What is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?

OA [spoiler]12/35[/spoiler]
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by jimmiejaz » Wed Nov 26, 2008 1:40 am
Total no of coins = 15
Total no of coins excluding nickels = 15-6=9
So, probability
= 9/15 * 8/14
= 12/35
What if i have not yet beat the beast, I know i will beat it!!!!!!!!
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by vivek.kapoor83 » Wed Nov 26, 2008 2:35 am
pls help me where i am getting wrong

prob of picking 1st nickel = 6/15
prob of picking 2nd nickel = 5/14

Prob of not picking 2 nickels in a row = 1-(6/15*5/14) =6/7
Where i am getting wrong pls explain
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by mals24 » Wed Nov 26, 2008 5:06 am
Where i am getting wrong pls explain
The error in your reasoning is that the negation of None is alteast and not all.

So the negation of getting no nickle will be getting atleast 1 nickle and not getting nickles in both tries.

So formula = 1 - P(atleast 1 nickle)

P(atleast 1 nickle) = (6*9)/(15*14)+(6*5)/(15*14)+(6*9)/(15*14) = 23/35

(6*9)/(15*14) = prob. of getting nickle in the first try and dime or pennies in the 2nd.

(6*9)/(15*14) = prob. of getting nickle in the 2nd try and dime or penny in the 1st

(6*5)/(15*14) = prob. of getting nickle in both tries.

Answer = 1-23/35 = 12/35

Hope you understood the logic :)
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