Probability

[monie411]

I got the following problem wrong. I definitely understand the solution but I'm not sure I get why my method didn't work. Here it is:

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

I decided to solve it by taking the probability that I would select a nickel twice and subtracting the result from 1. I got:

6/15*5/14 = 30/210 or 1/7

then the solution would be 6/7 (which it obviously isn't).

Why didn't my method work?

[TP]

I got the following problem wrong. I definitely understand the solution but I'm not sure I get why my method didn't work. Here it is:

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

I decided to solve it by taking the probability that I would select a nickel twice and subtracting the result from 1. I got:

6/15*5/14 = 30/210 or 1/7

then the solution would be 6/7 (which it obviously isn't).

Why didn't my method work?

Doesn't your solution include the probability that one of the 2 coins can
be a nickel?

Your P = 2 nickel coins together
Your 1 - P = addition of the following:
1) Non-Nickel + Non-Nickel
2) Nickel + Non-Nickel
3) Non-Nickel + Nickel

Does this reasoning sound correct? (I am not sure of it myself)

[Guest]

I came across same problem and solved it similary. i.e. by calculating the probability of event happening and the answer came to 6/7. What is wrong in my approach ?

[Sam]

I haven't come across the question, but assuming that 12/35 is the answer, try the followig:

If you want to approach it your way, it would be:

All the possible outcomes --> P(Nickel, then Nickel) + P(Non-Nickel, then Nickel) + P(Nickel, then Non-Nickel) + P(Non-Nickel, then Non-Nickel) = 1

What we want is:
P(Non-Nickel, then Non-Nickel) or

your preferred way which would be:
1 - [ P(Nickel, then Nickel) + P(Non-Nickel, then Nickel) + P(Nickel, then Non-Nickel) ]

1 - [ (6/15 x 5/14) + (6/15 x 9/14) + (9/15 x 6/14) ]

Which would reduce to

1 - 23/25 = 12/25

Is that the answer?

[Guest]

I completely agree with you Sam, I did not consider all the permutations of selecting Nickel. I just considered 1 case which was mentioned in the question. Yes 12/35 is the answer. Thank you very much for your response !

[jwinawer]

Nice work. This is what we call the (1-x) method for probability. Sometimes when you have to calculate the probablity of X, it is actually easier to calculate not X and then subtract from 1. But of course you need to be especially careful that the two situations you consider are the ONLY possible situations, otherwise their probabilities do not sum to 1, and the method is invalid, as pointed out above.

[chuddy09]

I am using C! to solve this issue

There are 6 N, 5 P and 4 D. In order to not select nickle twice in a row, we would only be selecting P and D -without replacement. Hence the formula should be

(5c1/9c1)X (4c1x8c1)

This gives me an answer of 5/18. What am I doing wrong? Plz advice.

[RonPurewal]

whoa, guys

no reason to use any sort of indirect solution (like the "1 - x" sort of thing) on this problem.
indirect solutions are best suited to problems on which the ACTUAL event under consideration is a COMPLEX EVENT - i.e., it comprises many different possibilities, and would thus be tedious/laborious to calculate directly.
for instance, if you get
what is the probability of getting at least one head on five tosses of a coin?"
then that's a COMPLEX EVENT, so you invoke the "1 - x" trick to shorten the job.

--

this is just about the world's worst problem for the "1 - x" method, since:
* this is NOT a complex event;
* the opposite event IS a complex event (!)

this is a basic problem in consecutive probability; you shouldn't even be considering methods like "1 - x" here.

the purse contains 6 nickels and 9 non-nickels.
therefore, the chance of selecting a non-nickel on the first try is 9/15, which reduces to 3/5.
after that, there are 6 nickels and 8 non-nickels left, so the chance of selecting another non-nickel is 8/14, or 4/7.
multiply them: 3/5 x 4/7 = 12/35.
done.

that is all.

@ previous poster:

I am using C! to solve this issue

don't.
this problem is nowhere near complicated enough to merit the use of "c" formulas.

why would you create the unnecessary work for yourself?
e.g., if i say "what's the probability that a randomly chosen digit is odd?" i would hope that you would just respond with 5 out of 10, and not "5c1/10c1".
if your response would be the latter, then you're probably going to be spending 3 minutes solving problems that are meant to take 30 seconds. on some other tests, that wouldn't be a big deal, but, on a test like this one, that's life and death.

[jeremysage26]

I need clarification on this problem. On a similar problem you needed to multiply the results by 2 because you could select penny then dime, or dime then penny. Why does this not apply here? I'm confused. Thank you in advance.

[tim]

you could do this if you wanted to complicate the problem by considering all the cases. you could calculate penny then dime as well as dime then penny, but then you'd also need to calculate dime then dime and penny then penny. since the problem does not require one of each, this is fundamentally different from the problem type you're referring to..