Die problem hint needed

[Student443]

A regular die was rolled 9 times. Calculate the probability that at least one six was drawn and 1 points appeared exactly once.

I have a problem with combining the both conditions together, since I know that ,,at least one six drawn" can be done by complementary events and ,,1 point exactly once" perhaps with Bernoulli trial (n=9,k=1), also know that omega = 6^9. Can anybody help me how to combine both conditions in one equation, that's the last step I miss?

 

[R.M.]

I'm a bit confused on the problem's wording, specifically "and 1 points appeared exactly once."

If I'm interpreting the problem correctly, it's asking for the probability that in 9 rolls of a die, "6" came up at least once, and "1" came up exactly once. Is that correct?

 

[Student443]

I'm a bit confused on the problem's wording, specifically "and 1 points appeared exactly once."

If I'm interpreting the problem correctly, it's asking for the probability that in 9 rolls of a die, "6" came up at least once, and "1" came up exactly once. Is that correct?

Yes, correct. I should of use one convention for ,,6" and ,,1", sorry for that.

 

[R.M.]

I'm not an expert on probability, so my method will probably be considered the "long way", but I can offer some thoughts on how I would attack it:
1. Total ways to roll a die nine times: 6⁹.
2. Imagine each die roll is a "slot", and also assume that exactly one 6 is rolled and one 1 is rolled. Let's also assume that the 6 is in the 1st slot, and the 1 is in the 2nd slot. Here are the number of possibilities: 1 x 1 x 4 x 4 x 4 x 4 x 4 x 4 x 4. But, the 6 can be anywhere and so could the 1, so we need to shuffle them around. ₉C₁ shuffles the 6, and ₈C₁ shuffles the 1 (they can't be in the same spot at the same time). 4⁷ x ₉C₁ x ₈C₁
3. Now let's assume that exactly two 6's are rolled and one 1 is rolled. Let's also assume that the 6 is in the 1st two slots, and the 1 is in the 3rd slot. Here are the number of possibilities: 1 x 1 x 1 x 4 x 4 x 4 x 4 x 4 x 4. Again, we need to shuffle the 6 and 1 around. ₉C₂ shuffles the 6, and ₇C₁ shuffles the 1. 4⁶ x ₉C₂ x ₇C₁
4. Notice a pattern forming: 4⁷ x ₉C₁ x ₈C₁ + 4⁶ x ₉C₂ x ₇C₁ + ...
5. Once you have that sum, divide it by the answer found in 1, and you should have your probability.

 

[pka]

A regular die was rolled 9 times. Calculate the probability that at least one six was drawn and 1 points appeared exactly once.
I have a problem with combining the both conditions together, since I know that ,,at least one six drawn" can be done by complementary events and ,,1 point exactly once" perhaps with Bernoulli trial (n=9,k=1), also know that omega = 6^9. Can anybody help me how to combine both conditions in one equation, that's the last step I miss?

First let's look one case. The case of having one \(1\), three sixes and five other numbers: \(1666XXXXX\). There each \(X\in\{2,3,4,5\}\).
That single can be arranged in \(\dfrac{9!}{(3!(5!))}\cdot 4^5\) ways.
So there are \(\displaystyle\sum\limits_{k = 1}^7 {\frac{{9!}}{{(8 - k)!k!}}{{(4)}^{8 - k}}} \) ordered \(9\)-tuples that meet the requirements.
To find the probability SEE HERE