Probability of not getting a prize

[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 $p(not ~ W) = \frac { {}^{9990}C_2} { {}^{10000}C_2}$

 

[mjc123]

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

 

[Hornbein]

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.

 

[erobz]

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

 

[vcsharp2003]

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".
$$p(not W) =\dfrac {{}^{3}C_2}{{}^{4}C_2} $$

 

[Hornbein]

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.

 

[vcsharp2003]

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?

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

 

[vcsharp2003]

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.

 

[erobz]

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:

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

 

[vcsharp2003]

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 $\dfrac {1}{2}$.

 

[erobz]

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.

 

[vcsharp2003]

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.

 

[vcsharp2003]

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?

 

[erobz]

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.

 

[erobz]

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]

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.

 

[vcsharp2003]

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.

 

[erobz]

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.

 

[vcsharp2003]

( ticket 1 or ticket 2 can be the winner)

But wouldn't doing that duplicate the number of ways of winning only 1 ticket?

 

[erobz]

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.

 

[haruspex]

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]

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.

 

[vcsharp2003]

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.

 

[haruspex]

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?

 

[erobz]

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]

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.

 

[haruspex]

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.

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

Right. So what is the probability of that?

 

[vcsharp2003]

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.

 

[erobz]

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.

 

[vcsharp2003]

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?

Right. So what is the probability of that?

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

 

[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.