There are 6 weights of 1 unit, 3 weights of 2 units, and 1 empty bag. Using one die, the following trial is performed. (Here, the unit of weight is g.) Roll the die once. If the number shown is 2 or less, put one weight of 1 unit into the bag. If the number shown is 3 or more, put one weight of 2 units into the bag. Repeat this trial until the total weight of the weights in the bag is first greater than or equal to 6. Let $X$ be the random variable representing the number of weights in the bag. The following is the process of finding the probability mass function $\mathrm { P } ( X = x ) ( x = 3,4,5,6 )$ of $X$. (i) The event $X = 3$ is the case where 3 weights of 2 units are in the bag, so $$\mathrm { P } ( X = 3 ) = \text{ (a) }$$ (ii) The event $X = 4$ can be divided into the case where the total weight of weights put in by the third trial is 4 and a weight of 2 units is put in on the fourth trial, and the case where the total weight of weights put in by the third trial is 5. Therefore, $$\mathrm { P } ( X = 4 ) = \left( \text{ (b) } + { } _ { 3 } \mathrm { C } _ { 1 } \left( \frac { 1 } { 3 } \right) ^ { 1 } \left( \frac { 2 } { 3 } \right) ^ { 2 } \right) \times \frac { 2 } { 3 }$$ (iii) The event $X = 5$ can be divided into the case where the total weight of weights put in by the fourth trial is 4 and a weight of 2 units is put in on the fifth trial, and the case where the total weight of weights put in by the fourth trial is 5. Therefore, $$\mathrm { P } ( X = 5 ) = { } _ { 4 } \mathrm { C } _ { 4 } \left( \frac { 1 } { 3 } \right) ^ { 4 } \left( \frac { 2 } { 3 } \right) ^ { 0 } \times \frac { 2 } { 3 } + \text{ (c) }$$ (iv) The event $X = 6$ is the case where the total weight of weights put in by the fifth trial is 5, so $$\mathrm { P } ( X = 6 ) = \left( \frac { 1 } { 3 } \right) ^ { 5 }$$ If the values corresponding to (a), (b), (c) are $a , b , c$ respectively, what is the value of $\frac { a b } { c }$? [4 points] (1) $\frac { 4 } { 9 }$ (2) $\frac { 7 } { 9 }$ (3) $\frac { 10 } { 9 }$ (4) $\frac { 13 } { 9 }$ (5) $\frac { 16 } { 9 }$
There are 6 weights of 1 unit, 3 weights of 2 units, and 1 empty bag. Using one die, the following trial is performed. (Here, the unit of weight is g.)
Roll the die once. If the number shown is 2 or less, put one weight of 1 unit into the bag. If the number shown is 3 or more, put one weight of 2 units into the bag.
Repeat this trial until the total weight of the weights in the bag is first greater than or equal to 6. Let $X$ be the random variable representing the number of weights in the bag. The following is the process of finding the probability mass function $\mathrm { P } ( X = x ) ( x = 3,4,5,6 )$ of $X$.\\
(i) The event $X = 3$ is the case where 3 weights of 2 units are in the bag, so
$$\mathrm { P } ( X = 3 ) = \text{ (a) }$$
(ii) The event $X = 4$ can be divided into the case where the total weight of weights put in by the third trial is 4 and a weight of 2 units is put in on the fourth trial, and the case where the total weight of weights put in by the third trial is 5. Therefore,
$$\mathrm { P } ( X = 4 ) = \left( \text{ (b) } + { } _ { 3 } \mathrm { C } _ { 1 } \left( \frac { 1 } { 3 } \right) ^ { 1 } \left( \frac { 2 } { 3 } \right) ^ { 2 } \right) \times \frac { 2 } { 3 }$$
(iii) The event $X = 5$ can be divided into the case where the total weight of weights put in by the fourth trial is 4 and a weight of 2 units is put in on the fifth trial, and the case where the total weight of weights put in by the fourth trial is 5. Therefore,
$$\mathrm { P } ( X = 5 ) = { } _ { 4 } \mathrm { C } _ { 4 } \left( \frac { 1 } { 3 } \right) ^ { 4 } \left( \frac { 2 } { 3 } \right) ^ { 0 } \times \frac { 2 } { 3 } + \text{ (c) }$$
(iv) The event $X = 6$ is the case where the total weight of weights put in by the fifth trial is 5, so
$$\mathrm { P } ( X = 6 ) = \left( \frac { 1 } { 3 } \right) ^ { 5 }$$
If the values corresponding to (a), (b), (c) are $a , b , c$ respectively, what is the value of $\frac { a b } { c }$? [4 points]\\
(1) $\frac { 4 } { 9 }$\\
(2) $\frac { 7 } { 9 }$\\
(3) $\frac { 10 } { 9 }$\\
(4) $\frac { 13 } { 9 }$\\
(5) $\frac { 16 } { 9 }$