Students are doing an experiment: that two fair dice are tossed, and all the outcomes are recorded. Then, an event is described as:the sum of the two up faces are recorded. Describe the random variable, X, in this event. Writethe sample space of this event. Calculatethe probability of rolling a 4, and probability of rolling a 12.

1.The random variable X can take any value from( 2,- ,12) in the event of the sample space.

2.The sample space S = [(1, 1), (1, 2),(2, 1), (1, 3),  (2, 2),(3, 1),  (1, 4),(2, 3),(3, 2),(4, 1), (1, 5),(2, 4), (3, 3),(4, 2),(5, 1),(1, 6),  (2, 5),(3, 4),(4, 3),(5, 2), (6, 1),   (2, 6),(3, 5),   (4, 4), (5, 3), (6, 2),(3, 6), (4, 5),  (5, 4), (6, 3),(4, 6) (5, 5), (6, 4),(5,6),  (6,5),(6,6)]

3. The probability of getting a 4 =   1/12

4. The probability of getting a 12 = 1/36

Step-by-step explanation:

There are 36 outcomes in all and the event shows the possible numbers obtained after the sum .

The event will be E= [ 2,3,4,5,6,7,8,9,10,11,12]

The random variable X can take any value from( 2,- ,12) in the event of the sample space.

The sample space S = [(1, 1), (1, 2),(2, 1), (1, 3),  (2, 2),(3, 1),  (1, 4),(2, 3),(3, 2),(4, 1), (1, 5),(2, 4), (3, 3),(4, 2),(5, 1),(1, 6),  (2, 5),(3, 4),(4, 3),(5, 2), (6, 1),   (2, 6),(3, 5),   (4, 4), (5, 3), (6, 2),(3, 6), (4, 5),  (5, 4), (6, 3),(4, 6) (5, 5), (6, 4),(5,6),  (6,5),(6,6)]

that is it has all the possible outcomes of both the dice.

The probability of getting a 4= Number of fav outcome/ no of possible outcomes=  3/ 36= 1/12

The probability of getting a 12= Number of fav outcome/ no of possible outcomes=1/36

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

answer: f = 32.2m

step-by-step explanation: that is the answer