changhiskhan 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
There are
6 nickels = 6N
5 pennies = 5P
4 dimes D = 4D
What is the probability of picking
[any] a coin
other than that is different from a nickel twice in a row (no replacement)?
I do not think that this question is poorly written except for the extra bits about what each coin is worth. I do think its language might be colloquial. I've put some different language in to show what "a" and "other than" signify.
"Not nickel" = anything other than, or different from, a nickel = pennies and dimes.
Change "not N" to the "affirmative" case: yes to pennies and dimes. When defining "favorable outcome," both P and D are included. Both are different from N.
I think FCP is easiest.
There are 15 coins total. 5P + 4D = 9 coins that give me the desired outcome.
For first pick, total outcomes = 15. Desired outcomes = 9. Probability of desired outcome is \(\frac{9}{15}\): I have 9 chances out of 15 total to get a D or a P.
The total number of coins is now 14. And there is one fewer coin in the desired group. So for second pick, probability of desired outcome is \(\frac{8}{14}\)
\(\frac{9}{15}\) * \(\frac{8}{14}\) = \(\frac{12}{35}\)
Answer B