Question

i need help solving my first question

Answer 1

Sorry, but it's too small. :\

Answer 2

You have to make an event table

Answer 3

Or just write all the required outcomes

Answer 4

Can you do that?

Answer 5

123 134 145 156 234 245 255 345 356 456
124 135 146 235 246 346
125 136 236
126

Answer 6

Now count the number of events by the probability of one of them(since the probability of each one remains same)

Answer 7

Is the ans 0.556??

Answer 8

i dont know, im still trying to figure it out myself

Answer 9

I am getting 0.0926

Answer 10

20(1/6 x 1/6 x 1/6)

Answer 11

wellthats not even a choice

Answer 12

What are the choices

Answer 13

Don't u have an ans key to check the ans ??

Answer 14

do you mean 0.596

Answer 15

2nd option i meant in the very first question of yours ..

Answer 16

the answers are just under the question

Answer 17

I know that will u pls tell if the ans i got was right or not from whatever u have (an ans key) so that i may make u understand regarding the same ..

Answer 18

Oh wait I listed the wrong events

Answer 19

i dont have the answer key, this is my homework and i dont understand it

Answer 20

Okk !!
Then shall i tell you how i got this??

Answer 21

of course, lets hear it

Answer 22

Okay i begin then!
First of all we want that no similar number should cum on the die so we can say that like first case is
1,2,3
Now if i keep 1 to be fix on one die then you want that 1 should not repeat on the remaining die so u don't want that number so choice is 5 out of 6 nos. on second dice and 4 out of 6 for third dice..

Answer 23

Hmm yeah its correct

Answer 24

Total cases are 5*4=20 for the case that 1 is on first dice..
Now do it for assuming the case that 2 is on dice 1 and u again get 20 cases..
So this process is repeated till 6 no. on dice 1 so u get total cases to be 120(20*6)