A dodecahedral die has 12 faces, numbered 1 - 12. If the die is weighted in such a way that 2 is twice as likely to be facing up as 1, 3 is three times as likely to be facing up as 1, and so on, what is the probability that 4 will land facing up
A) $\frac { 1 } { 12 }$
B) $\frac { 2 } { 39 }$
C) $\frac { 2 } { 33 }$
D) $\frac { 1 } { 6 }$
E) $\frac { 17 } { 78 }$

Question

A dodecahedral die has 12 faces, numbered 1 - 12. If the die is weighted in such a way that 2 is twice as likely to be facing up as 1, 3 is three times as likely to be facing up as 1, and so on, what is the probability that 4 will land facing up   
A) $\frac { 1 } { 12 }$
B) $\frac { 2 } { 39 }$
C) $\frac { 2 } { 33 }$
D) $\frac { 1 } { 6 }$
E) $\frac { 17 } { 78 }$

Quizplus · Accepted Answer

The probability of each number facing up is proportional to its value, so the total probability is the sum of the first 12 positive integers (1+2+3+...+12), which equals $\frac{12 \times 13}{2} = 78$. The probability of 4 facing up is $4/78 = \frac{2}{39}$.

A Dodecahedral Die Has 12 Faces, Numbered 1 - 12 112\frac { 1 } { 12 }121​

A Dodecahedral Die Has 12 Faces, Numbered 1 - 12 $\frac { 1 } { 12 }$