Suppose That a Bag Holds 3 Blue Balls and One$E ( X ) = \sum _ { n = 1 } ^ { \infty } n \frac { 3 { n } } { 4 ^ { n + 1 } } = \frac { 1 } { 4 } \sum _ { n = 1 } ^ { \infty } n \left( \frac { 3 } { 4 } \right) ^ { n } = 3$

Question

Suppose That a Bag Holds 3 Blue Balls and One$E ( X ) = \sum _ { n = 1 } ^ { \infty } n \frac { 3  { n } } { 4 ^ { n + 1 } } = \frac { 1 } { 4 } \sum _ { n = 1 } ^ { \infty } n \left( \frac { 3 } { 4 } \right) ^ { n } = 3$

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The Answer of Suppose That a Bag Holds 3 Blue Balls and One$E ( X ) = \sum _ { n = 1 } ^ { \infty } n \frac { 3  { n } } { 4 ^ { n + 1 } } = \frac { 1 } { 4 } \sum _ { n = 1 } ^ { \infty } n \left( \frac { 3 } { 4 } \right) ^ { n } = 3$

Suppose That a Bag Holds 3 Blue Balls and One

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