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

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==Solution==
 
==Solution==
  
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The third die cannot be <math>1</math>, since the minimal sum on the other two dice is <math>2</math>.
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The third toss cannot be <math>1</math>, since the minimal sum on the other two tosses is <math>2</math>.
  
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If the third die is <math>2</math>, then the first two dice must be <math>(1,1)</math>.
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If the third roll is <math>2</math>, then the first two rolls must be <math>(1,1)</math>.
  
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If the third die is <math>3</math>, then the first two dice must be <math>(1,2)</math> or <math>(2,1)</math>.
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If the third roll is <math>3</math>, then the first two rolls must be <math>(1,2)</math> or <math>(2,1)</math>.
  
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If the third die is <math>4</math>, then the first two dice must be <math>(1,3)</math>, <math>(2,2)</math>, or <math>(3,1)</math>.
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If the third roll is <math>4</math>, then the first two rolls must be <math>(1,3)</math>, <math>(2,2)</math>, or <math>(3,1)</math>.
  
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If the third die is <math>5</math>, then the first two dice must be <math>(1,4)</math>, <math>(2,3)</math>, <math>(3,2)</math>, or <math>(4,1)</math>.
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If the third roll is <math>5</math>, then the first two rolls must be <math>(1,4)</math>, <math>(2,3)</math>, <math>(3,2)</math>, or <math>(4,1)</math>.
  
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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>.
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If the third roll is <math>6</math>, then the first two rolls must be <math>(1,5)</math>, <math>(2,4)</math>, <math>(3,3)</math>, <math>(4,2)</math>, or <math>(5,1)</math>
  
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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>.
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There are <math>15</math> possibilities for the three tosses.  Of those possibilities, <math>7</math> have a <math>2</math> in the first two rolls, and <math>1</math> has a <math>2</math> in the third.  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}}
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{{MAA Notice}}

Latest revision as of 10:06, 12 May 2021

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 toss cannot be $1$, since the minimal sum on the other two tosses is $2$.

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

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

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

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

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

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

See also

1996 AHSME (Problems • Answer Key • Resources)
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

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