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+0  
 
+1
140
3
avatar+928 

Answer: 7/15

 

I have no idea how to approach this.

 Jul 28, 2019
 #1
avatar+105686 
+3

They are not hard to count.

 

|a-b|=c

a and b are interchangeable so there are really twice as many oucomes in the sample space

So that is 30 possible outcomes

Count those with a 3 and double it as well.    7*2=14 favourable outcomes

 

Prob is 14/30 = 7/15

 

 

  a b c
  6 5 1
  5 4 1
3 4 3 1
3 3 2 1
  2 1 1
  6 4 2
3 5 3 2
  4 2 2
3 3 1 2
3 6 3 3
3 5 2 3
3 4 1 3
  6 2 4
  5 1 4
  6 1 5
 Jul 28, 2019
 #2
avatar+23339 
+4

A fair six-sided number cube with the digits 1-6 on its faces is rolled three times.
The positive difference between the values on the first two rolls is equal to the value of the third roll.
What is the probability that at least one 3 was rolled?
Express your answer as a common fraction.

 

\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 1} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(0) & \text{_ _ _ _ _ _} \\ & 2~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & {\color{red}3}~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 4~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 5~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & 6~(5) & \text{_ _ _ _ $\color{blue}x$ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{2} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 2~(0) & \text{_ _ _ _ _ _} \\ & {\color{red}3}~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 4~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & 5~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 6~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ {\color{red}3}} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 2~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & {\color{red}3}~(0) & \text{_ _ _ _ _ _} \\ & 4~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 5~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 6~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 4} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 2~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & {\color{red}3}~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 4~(0) & \text{_ _ _ _ _ _} \\ & 5~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 6~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 5} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & 2~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & {\color{red}3}~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 4~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 5~(0) & \text{_ _ _ _ _ _} \\ & 6~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 6} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(5) & \text{_ _ _ _ $\color{blue}x$ _} \\ & 2~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & {\color{red}3}~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 4~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & 5~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 6~(0) & \text{_ _ _ _ _ _} \\ \hline \end{array}\)

 

The probability that at least one 3 was rolled is \(\mathbf{ \dfrac{14\ {\color{red}x} }{30\ {\color{blue}x} } = \dfrac{7}{15} }\)

 

laugh

 Jul 28, 2019
edited by heureka  Jul 28, 2019
 #3
avatar+928 
+2

Thanks for all your help guys!!!

dgfgrafgdfge111  Jul 28, 2019

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