A student has five nickels and four quarters in her pocket. She picks two coins at random. What is the
expected total value in cents of the coins?
(a) 125/9 (b) 250/9 (c) 30 (d) 25/2 (e) 15 (f) none of the others
i did (5/9) * .05 + (4/9 *.25 and got A
this is wrong the answer is B
2006-10-13 09:20:38 · 4 answers · asked by Diggler AKA The Cab Driver 1 in Science & Mathematics ➔ Mathematics
Using the notation C(n,k) = n!/[k!(n-k)!], there are C(9,2) = 9!/[2!7!] = 36 ways to choose the 2 coins.
The possibilities are:
1. 2 nickels, which can be done in C(5,2)=5!/[2!3!] = 10 ways, so the probability is 10/36 = 5/18.
2. 2 quarters, and since C(4,2) = 4!/[2!2!] = 6, the probability is 6/36 = 1/6
3. 1 quarter and 1 dime, with probability 1-1/6-5/18 = 5/9.
Therefore, the expected value is
(5/9)*0.30 + (1/6)*0.5 + (5/18)*0.1 = 250/9.
2006-10-13 09:31:50 · answer #1 · answered by James L 5 · 2⤊ 0⤋
(5/9*.05 + 4/9*25)*2 because you are picking two coins, not one.
2006-10-13 09:28:35 · answer #2 · answered by victorschool1 5 · 0⤊ 0⤋
What you got (answer A) was the expected value of one coin, not a draw of two coins. Merely multiply your expected value of one coin by two to get the correct answer.
2006-10-13 11:12:27 · answer #3 · answered by Anonymous · 0⤊ 0⤋
right.....
2006-10-13 09:22:42 · answer #4 · answered by kkkkkkkkkkevin 1 · 0⤊ 1⤋