Is there a clear-cut formula for this scenario where there is no replacement?
更新1:
In any order, sorry! So it’s combination, not permutation
Marbles without replacement question?
2020-08-21 4:40 pm
If I have 2 blue marbles, 3 red marbles, and 5 yellow marbles, what is the probabilty of selecting 2 red marbles and 3 yellow marbles if I draw 5 times without replacement?
Is there a clear-cut formula for this scenario where there is no replacement?
Is there a clear-cut formula for this scenario where there is no replacement?
回答 (2)
2020-08-21 7:39 pm
Number of ways to select 5 marbles from 10 = ₁₀C₅
Number of ways to select 2 red marbles and 3 yellow marbles = ₃C₂ × ₅C₃
P(select 2 red marbles and 3 yellow marbles)
= ₃C₂ × ₅C₃ / ₁₀C₅
= 3 × 10 / 252
= 5/42
Number of ways to select 2 red marbles and 3 yellow marbles = ₃C₂ × ₅C₃
P(select 2 red marbles and 3 yellow marbles)
= ₃C₂ × ₅C₃ / ₁₀C₅
= 3 × 10 / 252
= 5/42
2020-08-21 5:08 pm
I will assume that the marbles are non-distinct, i.e. the order does not matter.
Let's assume that the marbles are selected in the order R, R, Y, Y, Y
so, 3/10 x 2/9 x 5/8 x 4/7 x 3/6 => 1/84
Now, the combination of 2 reds and 3 yellows can occur in 5C2 ways
Now, 5C2 = 5!/2!3! => 10
so, 10 x (1/84) = 10/84 = 5/42
:)>
Let's assume that the marbles are selected in the order R, R, Y, Y, Y
so, 3/10 x 2/9 x 5/8 x 4/7 x 3/6 => 1/84
Now, the combination of 2 reds and 3 yellows can occur in 5C2 ways
Now, 5C2 = 5!/2!3! => 10
so, 10 x (1/84) = 10/84 = 5/42
:)>
收錄日期: 2021-04-24 07:59:00
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https://hk.answers.yahoo.com/question/index?qid=20200821084038AAbVURF