# Probability of not getting a prize

• vcsharp2003
In summary, the conversation discusses the probability of winning a prize when purchasing two tickets from a pool of 10,000 tickets, where there are only 10 winning tickets. It is determined that the probability of not winning is equal to the probability of not winning on the first ticket multiplied by the probability of not winning on the second ticket, which is equal to 1/2. This is also equivalent to the probability of not winning being equal to 1 minus the probability of winning at least one ticket, which is also 1/2. There is also discussion about the original approach used, which did not account for the multiple ways to win on two tickets, and the revised approach that does take this into account.

#### vcsharp2003

Homework Statement
The question is part (b) as shown in screenshot.
Relevant Equations
##p(A) + p(not ~ A) = 1## Let W be the event that a prize is received. Then ##p(W) + p(not ~ W) = 1##. We need to find ##p(not W)## and so let's try to find ##p(W)## and then we can subtract it from 1 to get ##p(not ~ W)##.
The experiment is buying 2 tickets. So, $$p(W) = \frac { {}^{10}C_2} { {}^{10000}C_2}$$
Thus $$p(not ~ W) = 1 - p(W)$$ $$p(not ~ W) = 1 - \frac { {}^{10}C_2} { {}^{10000}C_2}$$

The logic used is that we select 2 tickets from 10 winning tickets i.e. ##{}^{10}C_2##, which is the number of ways that 2 tickets can win. Also the total number of ways of buying 2 tickets from 10,000 tickets is ##{}^{10000}C_2##.

But the answer in the back is given as ##p(not ~ W) = \frac { {}^{9990}C_2} { {}^{10000}C_2}##

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P(W) is not the probability of getting 2 winning tickets. It is the probability of getting at least 1 winning ticket.

• vcsharp2003
mjc123 said:
P(W) is not the probability of getting 2 winning tickets. It is the probability of getting at least 1 winning ticket.
And it is equal to one minus the chance of not winning anything.

• vcsharp2003
Imagine there were 4 tickets, with only 1 prize ticket. What is the probability you don't win purchasing two tickets?

• vcsharp2003
erobz said:
Imagine there were 4 tickets, with only 1 prize ticket. What is the probability you don't win purchasing two tickets?
So the experiment would be to purchase two tickets and the event is not winning that is called "not W".
$$p(not W) =\dfrac {{}^{3}C_2}{{}^{4}C_2}$$

When I was a teaching assistant in probability I taught the students to always try this trick first. Quite often it is easier to calculate the chance of something not happening.

• hutchphd and vcsharp2003
mjc123 said:
P(W) is not the probability of getting 2 winning tickets. It is the probability of getting at least 1 winning ticket.
So if I added ##{}^{10}C_1 \times {}^{9990}C_1## to the numerator of the probability that I have calculated, then that would be correct?

$$p(W) = \frac { {}^{10}C_2 + {}^{10}C_{1} \times {}^{9990}C_1} { {}^{10000}C_2}$$

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• erobz
Hornbein said:
When I was a teaching assistant in probability I taught the students to always try this trick first. Quite often it is easier to calculate the chance of something not happening.
That's what I tried, but didn't get the answer.

vcsharp2003 said:
So the experiment would be to purchase two tickets and the event is not winning that is called "not W".
$$p(not W) =\dfrac {{}^{3}C_2}{{}^{4}C_2}$$
You have half of the total tickets

$$P(W) = P( nW ) = \frac{1}{2}$$

Alternatively, The probability you not win on the first ticket is ## P(nW1) = \frac{3}{4}##. The probability you not win on the second ticket is ##P(nW2) = \frac{2}{3}##

The probability that you don't win buying two tickets is the probability of not winning on the first and the second ticket:

$$P(nW) = P(nW1) \cdot P(nW2) = \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2}$$

• Chenkel and vcsharp2003
erobz said:
You have half of the total tickets

$$P(W) = P( nW ) = \frac{1}{2}$$

Alternatively, The probability you not win on the first ticket is ## P(nW1) = \frac{3}{4}##. The probability you not win on the second ticket is ##P(nW2) = \frac{2}{3}##

The probability that you don't win buying two tickets is the probability of not winning on the first and the second ticket:

$$P(nW) = P(nW1) \cdot P(nW2) = \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2}$$
Even the probability that I found in post #5 comes to ##\dfrac {1}{2}##.

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vcsharp2003 said:
Even the probability that I found comes to ##\dfrac {1}{2}##.
Ahh, yes it does!

In your original approach you are choosing 2 winners, but there are multiple ways to win. You can win on ticket 1, ticket 2, or ticket 1 and 2. So you would need to account for the ways you could win.

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• vcsharp2003
erobz said:
You have half of the total tickets

$$P(W) = P( nW ) = \frac{1}{2}$$

Alternatively, The probability you not win on the first ticket is ## P(nW1) = \frac{3}{4}##. The probability you not win on the second ticket is ##P(nW2) = \frac{2}{3}##

The probability that you don't win buying two tickets is the probability of not winning on the first and the second ticket:

$$P(nW) = P(nW1) \cdot P(nW2) = \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2}$$
Is the second approach you've used considering two dependent events or two independent events? I think it's two dependent events i.e. p(nW1) and p(nW2) are two dependent events with the second one depending on the first one.

erobz said:
Ahh, yes it does!

In your original approach you are choosing 2 winners, but there are multiple ways to win. You can win on ticket 1, ticket 2, or ticket 1 and 2. So you would need to account for the ways you could win.
I have tried to do that in post#7. Does the new ##p(W)## look correct in post#7?

vcsharp2003 said:
Is the second approach you've used considering two dependent events or two independent events? I think it's two dependent events i.e. p(nW1) and p(nW2) are two dependent events with the second one depending on the first one.
Yeah, you not winning on the second ticket depends on you not winning on the first ticket. If you won on the first ticket, the probability you win on the second ticket is 0.

• vcsharp2003
vcsharp2003 said:
I have tried to do that in post#7. Does the new ##p(W)## look correct in post#7?
It don't think it's quite correct.

There are ##C(10,2) \cdot C(2,2)## ways to select 2 winners, so you have that first term right in the numerator.

The second term in the numerator needs some work. I believe you just have to realize there are ##2## boxes to choose from for the winner you select ( ticket 1 or ticket 2 can be the winner), and adjust that term accordingly.

• vcsharp2003
erobz said:
The second term in the numerator needs some work. I believe you just have to realize there are 2 boxes to choose from for the winner you select.
The way I have written the second term is first select the winning ticket and after that select a non-winning ticket; these two steps would yield the big step of winning only one ticket. Then I simply use the Fundamental Rule of Counting to get the number of ways of winning only one ticket.

erobz said:
There are C(10,2)⋅C(2,2) ways to select 2 winners, so you have that first term right in the numerator.
Yes, I get that.

vcsharp2003 said:
The way I have written the second term is first select the winning ticket and after that select a non-winning ticket; these two steps would yield the big step of winning only one ticket. Then I simply use the Fundamental Rule of Counting to get the number of ways of winning only one ticket.
The seconnd term should be ##C(10,1) \cdot C(2,1) \cdot C(9990,1)##

There are two configurations for each pair of selected tickets.

erobz said:
( ticket 1 or ticket 2 can be the winner)
But wouldn't doing that duplicate the number of ways of winning only 1 ticket?

Yeah, I think you are right. It would over select.

Sanity check with small numbers:

Imagine a bag of 6 marbles, 4 red, 2 green

there are 6 *5 = 30 ways to select 2 marbles

You can select both red in 4 *3 = 12 ways

You can select both green in 2*1 = 2 ways

And you can select 1 red and one green in C(2,1) 4*2 ways = 16 ways

$$12 + 2 + 16 = 30$$

$$P(1r,1g) = \frac{16}{30} = \frac{8}{15}$$

Using combinations, we would have:

$$P(1r,1g) = \frac{C(4,1) \cdot C(2,1)}{C(6,2)} = \frac{8}{15}$$

Thanks for keeping me in check!

Given that, ##P(W)## in #7 looks reasonable to me.

• vcsharp2003
vcsharp2003 said:
That's what I tried, but didn't get the answer.
No, you went the other way. You were asked for the probability of not winning but chose instead to find the probability of winning. That was a backward move.

• vcsharp2003
haruspex said:
No, you went the other way. You were asked for the probability of not winning but chose instead to find the probability of winning. That was a backward move.
I was going to use the fact that probability of not winning = 1 - probability of winning and that's why I started with probability of winning; finding the complement event's probability is a good approach only when it's easier to find.

erobz said:
Imagine a bag of 6 marbles, 4 red, 2 green

there are 6 *5 = 30 ways to select 2 marbles

You can select both red in 4 *3 = 12 ways

You can select both green in 2*1 = 2 ways
When I try to do your example, I am getting half the values that you have got.

Number of ways of selecting 2 marbles from a total of 4 red + 2 green marbles is ##{}^6C_2= 15##

Number of ways of selecting 2 red marbles from 4 red marbles = ##{}^4C_2= 6##

Number of ways of selecting 2 green marbles from 2 green marbles = ##{}^2C_2= 1##

I think what you found is permutations, in which order is important rather than combinations.

vcsharp2003 said:
I was going to use the fact that probability of not winning = 1 - probability of winning and that's why I started with probability of winning; finding the complement event's probability is a good approach only when it's easier to find.
How many ways are there of picking one losing ticket?
How many ways are there of picking two losing tickets?

• vcsharp2003
vcsharp2003 said:
When I try to do your example, I am getting half the values that you have got.

Number of ways of selecting 2 marbles from a total of 4 red + 2 green marbles is ##{}^6C_2= 15##

Number of ways of selecting 2 red marbles from 4 red marbles = ##{}^4C_2= 6##

Number of ways of selecting 2 green marbles from 2 green marbles = ##{}^2C_2= 1##

I think what you found is permutations, in which order is important rather than combinations.
I did it both ways?

• vcsharp2003
haruspex said:
How many ways are there of picking one losing ticket?
How many ways are there of picking two losing tickets?
Number of ways of picking 1 losing ticket = ## {}^{10}C_1 \times {}^{9990}C_1##

Number of ways of picking 2 losing tickets ## = {}^{9990}C_2##

Since there are 10 winning tickets out of a total of 10,000 tickets.

vcsharp2003 said:
Number of ways of picking 1 losing ticket = ## {}^{10}C_1 \times {}^{9990}C_1##
I didn’t mean one winning and one losing… just one losing.
vcsharp2003 said:
Number of ways of picking 2 losing tickets ## = {}^{9990}C_2##
Right. So what is the probability of that?

• vcsharp2003
erobz said:
I did it both ways?
Yes. I see that. But shouldn't combinations be the correct approach here since you're selecting and not arranging things. Or you were just trying to verify something.

vcsharp2003 said:
Yes. I see that. But shouldn't combinations be the correct approach here since you're selecting and not arranging things. Or you were just trying to verify something.
Either way will get you the appropriate probabilities as long as you are consistent in using combinations OR permutations throughout the computation.

haruspex said:
I didn’t mean one winning and one losing… just one losing.
But the random experiment is to buy 2 tickets. So if 1 ticket is the losing ticket then the other should be the winning ticket. Or not?
haruspex said:
Right. So what is the probability of that?
##p(2 ~ losing ~ tickets) =\dfrac { {}^{9990}C_2 } { {}^{10000}C_2}##

erobz said:
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.

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vcsharp2003 said:
But the random experiment is to buy 2 tickets.
Sure, but I was asking that as a step to the second question.
vcsharp2003 said:
##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
haruspex said:
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##.
haruspex said:
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.

vcsharp2003 said:
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.

erobz said:
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.

• vcsharp2003