Exercise 4.1 #3
A die is rolled twice. What is the probability that the sum of the faces is greater than 7, given that
(a) the first outcome was a 4?
We will let A = sum of faces greater than 7, and B = first outcome is 4.
We will use the following formula to find the probability \(P(A|B)=\frac { P({ A }\cap B) }{ P(B) }\)
If our first outcome is a 4, \(P(B)=\frac { 1 }{ 6 }\) and the only possibility for the sum to be greater than 7 is if we roll a 4, 5 or 6, meaning \(P(A\cap B)=\frac { 3 }{ 6 } \times \frac { 1 }{ 6 } =\frac { 3 }{ 36 }\)
Thus,
\(P(A|B)=\frac { \frac { 3 }{ 36 } }{ \frac { 1 }{ 6 } } =\frac { 3 }{ 36 } \times \frac { 6 }{ 1 } =\frac { 1 }{ 2 }\)
(b) the first outcome was greater than 3?
A = sum of faces greater than 7, and B = first outcome is greater than 3.
If our first outcome is greater than 3, then it could be either a 4, 5 or 6, meaning \(P(B)=\frac { 3 }{ 6 }\)
For the sum to be greater than 7 when we roll a 4, our second roll would have to be either a 4, 5 or 6, when we roll a 5, the second roll would have to be a 3, 4, 5 or 6. Finally if we roll a 6, the second roll would have to be a 2, 3, 4, 5 or 6, meaning \(P(A\cap B)=(\frac { 3 }{ 6 } \times \frac { 1 }{ 6 } )+(\frac { 4 }{ 6 } \times \frac { 1 }{ 6 } )+(\frac { 5 }{ 6 } \times \frac { 1 }{ 6 } )=\frac { 3 }{ 36 } +\frac { 4 }{ 36 } +\frac { 5 }{ 36 } =\frac { 12 }{ 36 }\)
Thus,
\(P(A|B)=\frac { \frac { 12 }{ 36 } }{ \frac { 3 }{ 6 } } =\frac { 12 }{ 36 } \times \frac { 6 }{ 3 } =\frac { 72 }{ 108 } =\frac { 2 }{ 3 }\)
(c) the first outcome was a 1?
A = sum of faces greater than 7, and B = first outcome is 1.
If our first outcome is a 1, \(P(B)=\frac { 1 }{ 6 }\) and there are no possibilities for the sum to be greater than 7. But letโs compute the formula and see what we get.
\(P(A\cap B)=\frac { 0 }{ 6 } \times \frac { 1 }{ 6 } =0\)
Thus,
\(P(A|B)=\frac { 0 }{ \frac { 1 }{ 6 } } =0\times \frac { 6 }{ 1 } =0\)
(d) the first outcome was less than 5?
A = sum of faces greater than 7, and B = first outcome is less than 5.
If our first outcome is less than 5, then it could be either a 4, 3, 2 or 1, meaning \(P(B)=\frac { 4 }{ 6 }\)
For the sum to be greater than 7 when we roll a 4, our second roll would have to be either a 4, 5 or 6, when we roll a 3, the second roll would have to be a 5 or 6, if we roll a 2, our second roll would have to be a 6, and if we roll a 1, there are no possibilities for the sum to be greater than 7. This means \(P(A\cap B)=(\frac { 3 }{ 6 } \times \frac { 1 }{ 6 } )+(\frac { 2 }{ 6 } \times \frac { 1 }{ 6 } )+(\frac { 1 }{ 6 } \times \frac { 1 }{ 6 } )+(\frac { 0 }{ 6 } \times \frac { 1 }{ 6 } )=\frac { 3 }{ 36 } +\frac { 2 }{ 36 } +\frac { 1 }{ 36 } +0=\frac { 6 }{ 36 }\)
Thus,
\(P(A|B)=\frac { \frac { 6 }{ 36 } }{ \frac { 4 }{ 6 } } =\frac { 6 }{ 36 } \times \frac { 6 }{ 4 } =\frac { 36 }{ 144 } =\frac { 1 }{ 4 }\)