An ordinary fair die is a cube with the numbers one through six on the sides represented by painted that. Imagine that such a die is rolled twice in succession and that the face values of the two goals are added together. This song is recorded as the outcome of a single trial of a random experiment. Compute the probability of each of the following events. Event a: the sun is greater than six

ok so what i think your trying to ask is if we roll two dice that the sum will be more then 6

Two dice

Assuming that the dice are unbiased or not " loaded".

Each side has the same probability, is 1/6 =0.16667, to turn up when rolled, if the die (D) is unbiased. The probability of a side turning up on D1 when 2 dice ( D1,D2) are rolled, is independent of the side turning up in D2. So this is an independent event.

How many ways can one get a sum total of 6 if D1 &D2 are rolled at the same time?

These are the possibilities

Case 1.

D1 =1 & D2=5

Or

D1= 5 & D2=1

Case 2.

D1 =2 & D2=4

Or

D1= 4 & D2= 2

Case 3. D1=3, D2=3

P3 =0.027778

Let's say, P 1 the probability for case 1 and P2 for case 2. There are no other cases.

The final probability P and is the sum total P = P1 + P2 + P3 the probability law of mutually exclusive events.

P1= 0.02778+ 0.02778 =0.055558

P2= 0.02778+0.02778 =0.055558

Same way,

P3=0.027778, when there is only one way to get the sum 6.

So, P = 0.138894

Based on truncating at the sixth decimal place.

A visual representation with two unbiased dice and the possible cases would also give the same result and is a short cut method. I like to derive from the basics.

Hope This Helps!!!

answer: idk

step-by-step explanation:

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92