Question

given that all three faces given that all three faces are different in a throw of three dice find the probability that at least one is a 6
(questions 2)the sum is 9​

Answer 1

Answer:

sorry I didn't know the answer

Answer 2

Step-by-step explanation:

If, I will throw 3 dice how mana outcomes are there??

If, I throw 1 dice there was a outcomes are=6

If, I throw 2 dice there was a outcomes=6×6=36

Like I will consilor first dice comes = 1,2,3,4,5,6

second dice comes 1,2,3,4,5,6

third dice comes 1,2,3,4,5,6

so there are three times six possible values =6,6,6

If I see total number values if there is 1 on first dice then 6 comes in second dice and 3 dice is also come 6 number values if there is 1 on first dice then 6 come in second dice and 3 dice is also come 6 num so doing like this there total num of values is multiplication

so these values are =6×6×6=216

total number of cases these are all possible outcomes so we find probably of getting a total of at least 6 at least 6 means 6 sum or more than 6 so see what does I do in this case find the 6 sum to take depletd removal like 1 deplet =(2,2,2) this is a deplet whose sum comes 6 we want 6 more coming or become so see what I will do to find this probably some at least comes 6 so I will find that sum It is saying at least 6 or greatet than 6 comes do I will do that probably so sum is less then 6 if sum is less than 6 suppose probably is = P1 probably P1 is 1 - P some less than cases are minimum sum is = (1+1+1) will be the minimum sum 3 less that no sum so sum less than 6 probably are less than 6 are (1,1,1),(1,1,2),(1,2,1),(3,1,1),(1,2,3),(2,2,1),(2,2,2),(3,1,1,),(1,3,1) so there is no case there is sum comes 5 if 3 will make more num so that become 6 so these are all case so these all casea are 6 less than 6 so these are total 10

so probably is = sum is less than 6 is to

19 by 216

so my probably of sum is (>6)

1 - is probably of sum = 1 - 19/216

226/213

106/108 answer