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
My approach was calculating what the probability of nickel getting picked twice in a row is and subtracting that probability by 1. Probability of a nickel getting picked the first time is 2/5 and second time is 5/14. (2/5)*(5/14) = 1/7. So I was expecting 6/7 to show up but that choice is not given. I understand the OA and its methodology but what's wrong with my approach?
Thanks!
In this case it would be easier to calculate the probability in direct way:
\(\frac{non-nickel}{all}*\frac{non-nickel -1}{all-1}=\frac{9}{15}*\frac{8}{14}=\frac{12}{35}\).
The problem in your solution is that the opposite probability of two non-nickel coins in a row is not two nickel in a row, it's two nickels in a row
plus either of two coins from two is nickel, (so basically the probability of at least one nickel from two picks):
\(\frac{6}{15}*\frac{5}{14}+2*\frac{6}{15}*\frac{9}{14}=\frac{23}{35}\)
\(P=1-\frac{23}{35}=\frac{12}{35}\)
Answer: B.