Joe 發問於 科學及數學數學 · 1 十年前

中學_probability - Urgent Thanks a lot

2 identical dice are rolled. P(2 odd numbers)=4/25, P(1 even & 1 odd)=12/25, P(2 even numbers)=9/25.

Now, rolling both dice for 3 times. If 2 odd numbers=score 10. If 1 even, 1 odd=score 5. If 2 even numbers=score 2.

What is the probability that the total score of the 3 trials is less than 10 ?

Ans:(9/25 x 9/25 x 9/25) + (9/25 x 9/25 x 12/25) x3

Question: Why not ..............+ (9/25 x 9/25 x 12/25) x2 ?

Please explains.

更新:

It must get (2 even numbers 3 times) or (2 even numbers 2 times and 1 even 1 odd)

1st trial: [even,even] , [even,even] , [even,even]

2nd trial: [even,even] , [even,even] , [even,even]

3rd trial: [even,even] , [even,odd] , [odd,even]

更新 2:

e.g. If score is larger than 20 = (4/25x4/25x4/25) + [4/25x4/25x (1-4/25)] x 3

1st trial: [odd,odd] , [odd,odd] , [odd,odd] , [odd,odd]

2nd trial: [odd,odd] ,[odd,odd] , [odd,odd] , [odd,odd]

3rd trial: [odd,odd] , [eve,eve] , [odd,eve] , [eve,odd]

更新 3:

Why [even,odd] = [odd,even]?? E.g. By rolling a die, P(odd)=2/5 and P(even)=3/5. Now 2 such identical dice are rolled.

Ans: P(1 even, 1 odd) = (2/5 x 3/5) + (3/5 x 2/5) = 12/25

Question: if [even,odd] = [odd,even], then why P(1 even , 1 odd) not= (2/5 x 3/5) = 6/25 ??

1 個解答

評分
  • 匿名
    1 十年前
    最愛解答

    The cases are mutually exclusive, as that

    [even,odd] = [odd,even] since the dice are identical

    case1:

    [even,even] , [even,even] , [even,even]

    case2:

    [even,odd] , [even,even] , [even,even]

    case3:

    [even,even] , [even,odd] , [even,even]

    case4:

    [even,even] , [even,even] , [even,odd]

    in maths term:

    (9/25 x 9/25 x 9/25) + (9/25 x 9/25 x 12/25) x3!/2

    =(9/25 x 9/25 x 9/25) + (9/25 x 9/25 x 12/25) x3

還有問題嗎?立即提問即可得到解答。