Probability of not getting a prize

[vcsharp2003]

Either way will get you the appropriate probabilities as long as you are consistent in using combinations OR permutations throughout the computation.

I see. I have to figure that out. I thought only combinations was the right approach in this example.

 

[haruspex]

But the random experiment is to buy 2 tickets.

Sure, but I was asking that as a step to the second question.

$p(2 ~ losing ~ tickets) =\dfrac { {}^{9990}C_2 } { {}^{10000}C_2}$

Right.. isn't that what you were trying to show, or have I misunderstood your issue?

 

[vcsharp2003]

Sure, but I was asking that as a step to the second question.

Then the random experiment would be to buy 1 ticket. The number of ways for losing 1 ticket would then be ${}^{9990}C_1$.

Right.. isn't that what you were trying to show, or have I misunderstood your issue?

Yes, you're correct. I could have directly found the number of ways of 2 tickets being losing tickets in order to find the probability of not getting a prize. If even 1 ticket is a win out of 2 purchased tickets, the person would still get a prize.

Thankyou for the help. It's solved.

 

[erobz]

I see. I have to figure that out. I thought only combinations was the right approach in this example.

If you are going to reach into the bag pull three marbles, what the probability of getting 2 red 1 green marble?

Using Combinations:

$$ P(2r,1g) = \frac{C(4,2)C(2,1)}{C(6,3)} = \frac{3}{5}$$

Using Permutations:

$$ P(2r,1g) = \frac{ 4 \cdot 3 \cdot C(3,2) \cdot 2}{ 6 \cdot 5 \cdot 4} = \frac{3}{5}$$

Combinations are computationally superior...I guess. Both work.

 

[haruspex]

Combinations are computationally superior...I guess. Both work.

It depends whether order matters. In the present case, it makes no difference whether you pick a winner, then a loser, or the other way around. So for k selections, the factor k! appears in the denominator for both the number of successful combinations and the total number of combinations.

In a random walk with an absorbing barrier the order does matter. If one step away from the barrier, what is the probability of reaching it in three goes (with 50:50 at each step)?
You only need two of the three in the right direction: an evens chance.
But if the first step is in the right direction it doesn’t matter about the other two: so 5/8.

 

[erobz]

It depends whether order matters. In the present case, it makes no difference whether you pick a winner, then a loser, or the other way around. So for k selections, the factor k! appears in the denominator for both the number of successful combinations and the total number of combinations.

In a random walk with an absorbing barrier the order does matter. If one step away from the barrier, what is the probability of reaching it in three goes (with 50:50 at each step)?
You only need two of the three in the right direction: an evens chance.
But if the first step is in the right direction it doesn’t matter about the other two: so 5/8.

I guess I didn't realize that might not be the case for all classes of problems. I don't have any formal coursework in the subject.

 

[vcsharp2003]

If you are going to reach into the bag pull three marbles, what the probability of getting 2 red 1 green marble?

Using Combinations:

$$ P(2r,1g) = \frac{C(4,2)C(2,1)}{C(6,3)} = \frac{3}{5}$$

Using Permutations:

$$ P(2r,1g) = \frac{ 4 \cdot 3 \cdot C(3,2) \cdot 2}{ 6 \cdot 5 \cdot 4} = \frac{3}{5}$$

Combinations are computationally superior...I guess. Both work.

As pointed out by @haruspex, when probability is determined, a common factor appears in numerator as well as denominator when taking permutations in this example according to the identity below. That's why using either permutations or combinations to determine probability gives the same value in the example of selecting marbles.

$${}^nP_r =r! \times {}^nC_r $$