Difference between revisions of "1996 AHSME Problems/Problem 16"

(Solution)
Line 19: Line 19:
 
If the third die is <math>6</math>, then the first two dice must be <math>(1,5)</math>, <math>(2,4)</math>, <math>(3,3)</math>, <math>(4,2)</math>, or <math>(5,1)</math>.
 
If the third die is <math>6</math>, then the first two dice must be <math>(1,5)</math>, <math>(2,4)</math>, <math>(3,3)</math>, <math>(4,2)</math>, or <math>(5,1)</math>.
  
There are <math>15</math> possibilities for the three dice.  Of those possibiltiies, <math>7</math> have a <math>2</math> in the first two dice, and <math>1</math> has a <math>2</math> in the third die.  Therefore, the answer is <math>\boxed{8}{15}</math>.
+
There are <math>15</math> possibilities for the three dice.  Of those possibiltiies, <math>7</math> have a <math>2</math> in the first two dice, and <math>1</math> has a <math>2</math> in the third die.  Therefore, the answer is <math>\boxed{\frac{8}{15}}</math>.
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=15|num-a=17}}
 
{{AHSME box|year=1996|num-b=15|num-a=17}}

Revision as of 14:31, 19 August 2011

Problem

A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equal the third, what is the probability that at least one "2" is tossed?

$\text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{91}{216}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{8}{15}\qquad\text{(E)}\ \frac{7}{12}$

Solution

The third die cannot be $1$, since the minimal sum on the other two dice is $2$.

If the third die is $2$, then the first two dice must be $(1,1)$.

If the third die is $3$, then the first two dice must be $(1,2)$ or $(2,1)$.

If the third die is $4$, then the first two dice must be $(1,3)$, $(2,2)$, or $(3,1)$.

If the third die is $5$, then the first two dice must be $(1,4)$, $(2,3)$, $(3,2)$, or $(4,1)$.

If the third die is $6$, then the first two dice must be $(1,5)$, $(2,4)$, $(3,3)$, $(4,2)$, or $(5,1)$.

There are $15$ possibilities for the three dice. Of those possibiltiies, $7$ have a $2$ in the first two dice, and $1$ has a $2$ in the third die. Therefore, the answer is $\boxed{\frac{8}{15}}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions
Invalid username
Login to AoPS