Problem Solving

In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If 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?

a. 8/25
b. 12/35
c. 13/35
d. 9/25
e. 17/25


My approach to solving this problem was to take the probability of getting a nickel twice and subtracting one. Why doesn't that work here?

okigbo wrote:In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If 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?

a. 8/25
b. 12/35
c. 13/35
d. 9/25
e. 17/25


My approach to solving this problem was to take the probability of getting a nickel twice and subtracting one. Why doesn't that work here?

First draw is NOT a nickel and the second draw is NOT a nickel

= 9/15 × 8/14 = [spoiler]12/35[/spoiler].

[spoiler]B[/spoiler]

1 - the probability that the first two draws were nickels, is not same as NO nickel in the first two draws, that's why your idea didn't click.

Could you please elaborate - 1 - the probability that the first two draws were nickels, is not same as NO nickel in the first two draws, that's why your idea didn't click.


What am I missing ? When will the probability of not picking 2 nickel = 1 - (picking two nickels )

okigbo wrote: My approach to solving this problem was to take the probability of getting a nickel twice and subtracting one. Why doesn't that work here?

when you subtract a probability from 1, you are essentially finding out a complement. A complement is a set of all the possible cases which don't satisfy a particular condition.

In your method, you first found out probability of getting 2 nickles (6/7). You then found out its complement(1/7). Now the complement in your case means : "not having two nickles", which does not restrict you from having one nickel.

Therefore the complement will contain cases like a penny and a dime, a nickel and a penny, a nickel and a dime etc.

and this is not same as not having two nickles one after another.


Hence, the corret way would be as Sanju has done it.

Hope this is clear now!

yes. if u want to solve through compliments,
you have to account for each compliment of getting a nickel separately.
compliment of getting one nickel (6/15) * compliment of getting one nickel after one coin other than nickel has been picked(6/14) = [1-6/15]*[1-6/14] = 9/15 * 8/14 = 12/35.
since the question wants the case of NOT getting a nickel, 5/14 wont' work.
then you would be just finding the compliment of getting a nickel.
it tries to trick you right there IMO.
its best to find
the prob. of getting other than nickels since it's much simpler.