Probability of independent not mutually exclusive events

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I am wondering if I am doing this right. If an event has a probability of happening of $0.9%$, what is the probability it will happen if it is repeated $50$ times. I have tried to calculate this with this formula:

$1-0.009^50$

Which gives $1$, so what I am wondering is it really true that if an event has a chance of only $0.9%$ of happening once, if repeated $50$ times it will happen for sure, or have I used the wrong formula to calculate this?

probability combinatorics

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edited Jul 25 at 9:59

Jam

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asked Jul 25 at 9:43

Leff

1232

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You have calculated that the event does not happen 50 times in 50 trials.
  – callculus
  Jul 25 at 9:53
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

I am wondering if I am doing this right. If an event has a probability of happening of $0.9%$, what is the probability it will happen if it is repeated $50$ times. I have tried to calculate this with this formula:

$1-0.009^50$

Which gives $1$, so what I am wondering is it really true that if an event has a chance of only $0.9%$ of happening once, if repeated $50$ times it will happen for sure, or have I used the wrong formula to calculate this?

probability combinatorics

share|cite|improve this question

edited Jul 25 at 9:59

Jam

4,26111230

asked Jul 25 at 9:43

Leff

1232

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You have calculated that the event does not happen 50 times in 50 trials.
  – callculus
  Jul 25 at 9:53
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I am wondering if I am doing this right. If an event has a probability of happening of $0.9%$, what is the probability it will happen if it is repeated $50$ times. I have tried to calculate this with this formula:

$1-0.009^50$

Which gives $1$, so what I am wondering is it really true that if an event has a chance of only $0.9%$ of happening once, if repeated $50$ times it will happen for sure, or have I used the wrong formula to calculate this?

probability combinatorics

share|cite|improve this question

edited Jul 25 at 9:59

Jam

4,26111230

asked Jul 25 at 9:43

Leff

1232

I am wondering if I am doing this right. If an event has a probability of happening of $0.9%$, what is the probability it will happen if it is repeated $50$ times. I have tried to calculate this with this formula:

$1-0.009^50$

Which gives $1$, so what I am wondering is it really true that if an event has a chance of only $0.9%$ of happening once, if repeated $50$ times it will happen for sure, or have I used the wrong formula to calculate this?

probability combinatorics

share|cite|improve this question

edited Jul 25 at 9:59

Jam

4,26111230

asked Jul 25 at 9:43

Leff

1232

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 9:59

Jam

4,26111230

edited Jul 25 at 9:59

Jam

4,26111230

edited Jul 25 at 9:59

Jam

4,26111230

4,26111230

asked Jul 25 at 9:43

Leff

1232

asked Jul 25 at 9:43

Leff

1232

asked Jul 25 at 9:43

Leff

1232

1232

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You have calculated that the event does not happen 50 times in 50 trials.
  – callculus
  Jul 25 at 9:53
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You have calculated that the event does not happen 50 times in 50 trials.
  – callculus
  Jul 25 at 9:53
  

  
  
  
  

2

2

You have calculated that the event does not happen 50 times in 50 trials.
– callculus
Jul 25 at 9:53

You have calculated that the event does not happen 50 times in 50 trials.
– callculus
Jul 25 at 9:53

add a comment | 

4 Answers
4

active

oldest

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up vote
3
down vote



accepted

• $0.009^50$ is the probability the event happens $50$ times out of $50$


• $1-0.009^50$ is the probability the event happens fewer than $50$ times out of $50$


• $(1-0.009)^50$ is the probability the event happens $0$ times out of $50$


• $1-(1-0.009)^50$ is the probability the event happens more than $0$ times out of $50$

share|cite|improve this answer

edited Jul 25 at 10:10

answered Jul 25 at 9:49

Henry

92.9k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "the probability the event does not happens fewer than 50 times out of 50". Correct me if I'm wrong but isn't this the probability that the event does happen fewer than 50 times in 50 trials?
  – Jam
  Jul 25 at 9:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Jam - indeed - thank you - edited
  – Henry
  Jul 25 at 10:12
  

  
  
  
  

add a comment | 

up vote
2
down vote

The chance of happening is $0.009$. So what do you compute by
$$1-0.009^50 ?$$
That is the chance that in 50 tries, it happens at most 49 times, and of course that is quite high, given the small chance.

What you need to compute is
$$1 - 0.991^50 approx 36%,$$
I'll leave it to you to find out why. :)

share|cite|improve this answer

answered Jul 25 at 9:49

Dirk Liebhold

4,026117

add a comment | 

up vote
-1
down vote

Because your calculator can only calculate up to certain precision. For a value that is too small, say $0.009^5$ in your case, the calculator regards it as $0$.

share|cite|improve this answer

answered Jul 25 at 9:49

Jerry

326212

add a comment | 

up vote
-1
down vote

For starters 0.0009^50 is an extremely small amount and is so small that most calculators would regard it as 0 (this is because the product of any number less than one and another number less than one is smaller than the original number, for example 0.2*0.2 = 0.04 and 0.04<0.2) . As a result, 2-0.0009^50 is equal to 2.

0.9% is equal to 9/1000. So the probability of an event which occurs once with the probability of 0.9% is 1*9/1000. The probability of an event occurring 50 times with the probability of 0.9% is 50*9/1000 which equals 450/1000 and is simplified to $boxeddfrac9200$ or $boxed4.5 %$.

share|cite|improve this answer

edited Jul 25 at 13:52

answered Jul 25 at 10:00

Vivek

163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the probability of an event is $0.009$, then the probability it occurs $50$ times in $50$ trials is not $0.009times50$.
  – Jam
  Jul 25 at 10:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If that were the case, the probability it occurs $1$ million times in $1$ million trials would be $9000$, i.e. $900,000%$, which can't be right
  – Jam
  Jul 25 at 10:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If an event had the probability of occurring 0.9% of the time then in 1 million trials it would occur 0.009*1,000,000 times which is 9000 times. What I am trying to say (with your example) that it will occur 9000 times in one million trials. Because the probability of an event occurring is it's probability multiplied by the # of trials. I may not have been very clear with that.
  – Vivek
  Jul 25 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That works for finding the expected number of successes. And I agree that if you perform $n$ trials which are independent and each have a probability of success of $p$ then the expected number of successes is $np$. But the probability of success in all trials would be $P(t_1wedge t_2wedge ldots wedge t_n)=P(t_1)times P(t_2)ldots times P(t_n)=ldots$, where $P(t_i)$ is the probability of success in trial $i$. In your answer you've calculated an expected value, as if it were a probability
  – Jam
  Jul 25 at 13:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Intuitively, you can see that your answer must be flawed since you've got a higher probability when you have a success multiple times. Imagine if your probability of winning the lottery were $0.9%$. Would it not seem strange if your probability of winning the lottery $50$ times consecutively were $4.5%$?
  – Jam
  Jul 25 at 13:47
  
  

  
  
  
  

 | 
show 1 more comment

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

• $0.009^50$ is the probability the event happens $50$ times out of $50$


• $1-0.009^50$ is the probability the event happens fewer than $50$ times out of $50$


• $(1-0.009)^50$ is the probability the event happens $0$ times out of $50$


• $1-(1-0.009)^50$ is the probability the event happens more than $0$ times out of $50$

share|cite|improve this answer

edited Jul 25 at 10:10

answered Jul 25 at 9:49

Henry

92.9k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "the probability the event does not happens fewer than 50 times out of 50". Correct me if I'm wrong but isn't this the probability that the event does happen fewer than 50 times in 50 trials?
  – Jam
  Jul 25 at 9:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Jam - indeed - thank you - edited
  – Henry
  Jul 25 at 10:12
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

• $0.009^50$ is the probability the event happens $50$ times out of $50$


• $1-0.009^50$ is the probability the event happens fewer than $50$ times out of $50$


• $(1-0.009)^50$ is the probability the event happens $0$ times out of $50$


• $1-(1-0.009)^50$ is the probability the event happens more than $0$ times out of $50$

share|cite|improve this answer

edited Jul 25 at 10:10

answered Jul 25 at 9:49

Henry

92.9k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "the probability the event does not happens fewer than 50 times out of 50". Correct me if I'm wrong but isn't this the probability that the event does happen fewer than 50 times in 50 trials?
  – Jam
  Jul 25 at 9:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Jam - indeed - thank you - edited
  – Henry
  Jul 25 at 10:12
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

• $0.009^50$ is the probability the event happens $50$ times out of $50$


• $1-0.009^50$ is the probability the event happens fewer than $50$ times out of $50$


• $(1-0.009)^50$ is the probability the event happens $0$ times out of $50$


• $1-(1-0.009)^50$ is the probability the event happens more than $0$ times out of $50$

share|cite|improve this answer

edited Jul 25 at 10:10

answered Jul 25 at 9:49

Henry

92.9k469147

• $0.009^50$ is the probability the event happens $50$ times out of $50$


• $1-0.009^50$ is the probability the event happens fewer than $50$ times out of $50$


• $(1-0.009)^50$ is the probability the event happens $0$ times out of $50$


• $1-(1-0.009)^50$ is the probability the event happens more than $0$ times out of $50$

share|cite|improve this answer

edited Jul 25 at 10:10

answered Jul 25 at 9:49

Henry

92.9k469147

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 10:10

edited Jul 25 at 10:10

edited Jul 25 at 10:10

answered Jul 25 at 9:49

Henry

92.9k469147

answered Jul 25 at 9:49

Henry

92.9k469147

answered Jul 25 at 9:49

Henry

92.9k469147

92.9k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "the probability the event does not happens fewer than 50 times out of 50". Correct me if I'm wrong but isn't this the probability that the event does happen fewer than 50 times in 50 trials?
  – Jam
  Jul 25 at 9:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Jam - indeed - thank you - edited
  – Henry
  Jul 25 at 10:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "the probability the event does not happens fewer than 50 times out of 50". Correct me if I'm wrong but isn't this the probability that the event does happen fewer than 50 times in 50 trials?
  – Jam
  Jul 25 at 9:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Jam - indeed - thank you - edited
  – Henry
  Jul 25 at 10:12
  

  
  
  
  

"the probability the event does not happens fewer than 50 times out of 50". Correct me if I'm wrong but isn't this the probability that the event does happen fewer than 50 times in 50 trials?
– Jam
Jul 25 at 9:54

"the probability the event does not happens fewer than 50 times out of 50". Correct me if I'm wrong but isn't this the probability that the event does happen fewer than 50 times in 50 trials?
– Jam
Jul 25 at 9:54

1

1

@Jam - indeed - thank you - edited
– Henry
Jul 25 at 10:12

@Jam - indeed - thank you - edited
– Henry
Jul 25 at 10:12

add a comment | 

up vote
2
down vote

The chance of happening is $0.009$. So what do you compute by
$$1-0.009^50 ?$$
That is the chance that in 50 tries, it happens at most 49 times, and of course that is quite high, given the small chance.

What you need to compute is
$$1 - 0.991^50 approx 36%,$$
I'll leave it to you to find out why. :)

share|cite|improve this answer

answered Jul 25 at 9:49

Dirk Liebhold

4,026117

add a comment | 

up vote
2
down vote

The chance of happening is $0.009$. So what do you compute by
$$1-0.009^50 ?$$
That is the chance that in 50 tries, it happens at most 49 times, and of course that is quite high, given the small chance.

What you need to compute is
$$1 - 0.991^50 approx 36%,$$
I'll leave it to you to find out why. :)

share|cite|improve this answer

answered Jul 25 at 9:49

Dirk Liebhold

4,026117

add a comment | 

up vote
2
down vote

up vote
2
down vote

The chance of happening is $0.009$. So what do you compute by
$$1-0.009^50 ?$$
That is the chance that in 50 tries, it happens at most 49 times, and of course that is quite high, given the small chance.

What you need to compute is
$$1 - 0.991^50 approx 36%,$$
I'll leave it to you to find out why. :)

share|cite|improve this answer

answered Jul 25 at 9:49

Dirk Liebhold

4,026117

The chance of happening is $0.009$. So what do you compute by
$$1-0.009^50 ?$$
That is the chance that in 50 tries, it happens at most 49 times, and of course that is quite high, given the small chance.

What you need to compute is
$$1 - 0.991^50 approx 36%,$$
I'll leave it to you to find out why. :)

share|cite|improve this answer

answered Jul 25 at 9:49

Dirk Liebhold

4,026117

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 9:49

Dirk Liebhold

4,026117

answered Jul 25 at 9:49

Dirk Liebhold

4,026117

answered Jul 25 at 9:49

Dirk Liebhold

4,026117

4,026117

add a comment | 

add a comment | 

up vote
-1
down vote

Because your calculator can only calculate up to certain precision. For a value that is too small, say $0.009^5$ in your case, the calculator regards it as $0$.

share|cite|improve this answer

answered Jul 25 at 9:49

Jerry

326212

add a comment | 

up vote
-1
down vote

Because your calculator can only calculate up to certain precision. For a value that is too small, say $0.009^5$ in your case, the calculator regards it as $0$.

share|cite|improve this answer

answered Jul 25 at 9:49

Jerry

326212

add a comment | 

up vote
-1
down vote

up vote
-1
down vote

Because your calculator can only calculate up to certain precision. For a value that is too small, say $0.009^5$ in your case, the calculator regards it as $0$.

share|cite|improve this answer

answered Jul 25 at 9:49

Jerry

326212

Because your calculator can only calculate up to certain precision. For a value that is too small, say $0.009^5$ in your case, the calculator regards it as $0$.

share|cite|improve this answer

answered Jul 25 at 9:49

Jerry

326212

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 9:49

Jerry

326212

answered Jul 25 at 9:49

Jerry

326212

answered Jul 25 at 9:49

Jerry

326212

326212

add a comment | 

add a comment | 

up vote
-1
down vote

For starters 0.0009^50 is an extremely small amount and is so small that most calculators would regard it as 0 (this is because the product of any number less than one and another number less than one is smaller than the original number, for example 0.2*0.2 = 0.04 and 0.04<0.2) . As a result, 2-0.0009^50 is equal to 2.

0.9% is equal to 9/1000. So the probability of an event which occurs once with the probability of 0.9% is 1*9/1000. The probability of an event occurring 50 times with the probability of 0.9% is 50*9/1000 which equals 450/1000 and is simplified to $boxeddfrac9200$ or $boxed4.5 %$.

share|cite|improve this answer

edited Jul 25 at 13:52

answered Jul 25 at 10:00

Vivek

163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the probability of an event is $0.009$, then the probability it occurs $50$ times in $50$ trials is not $0.009times50$.
  – Jam
  Jul 25 at 10:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If that were the case, the probability it occurs $1$ million times in $1$ million trials would be $9000$, i.e. $900,000%$, which can't be right
  – Jam
  Jul 25 at 10:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If an event had the probability of occurring 0.9% of the time then in 1 million trials it would occur 0.009*1,000,000 times which is 9000 times. What I am trying to say (with your example) that it will occur 9000 times in one million trials. Because the probability of an event occurring is it's probability multiplied by the # of trials. I may not have been very clear with that.
  – Vivek
  Jul 25 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That works for finding the expected number of successes. And I agree that if you perform $n$ trials which are independent and each have a probability of success of $p$ then the expected number of successes is $np$. But the probability of success in all trials would be $P(t_1wedge t_2wedge ldots wedge t_n)=P(t_1)times P(t_2)ldots times P(t_n)=ldots$, where $P(t_i)$ is the probability of success in trial $i$. In your answer you've calculated an expected value, as if it were a probability
  – Jam
  Jul 25 at 13:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Intuitively, you can see that your answer must be flawed since you've got a higher probability when you have a success multiple times. Imagine if your probability of winning the lottery were $0.9%$. Would it not seem strange if your probability of winning the lottery $50$ times consecutively were $4.5%$?
  – Jam
  Jul 25 at 13:47
  
  

  
  
  
  

 | 
show 1 more comment

up vote
-1
down vote

For starters 0.0009^50 is an extremely small amount and is so small that most calculators would regard it as 0 (this is because the product of any number less than one and another number less than one is smaller than the original number, for example 0.2*0.2 = 0.04 and 0.04<0.2) . As a result, 2-0.0009^50 is equal to 2.

0.9% is equal to 9/1000. So the probability of an event which occurs once with the probability of 0.9% is 1*9/1000. The probability of an event occurring 50 times with the probability of 0.9% is 50*9/1000 which equals 450/1000 and is simplified to $boxeddfrac9200$ or $boxed4.5 %$.

share|cite|improve this answer

edited Jul 25 at 13:52

answered Jul 25 at 10:00

Vivek

163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the probability of an event is $0.009$, then the probability it occurs $50$ times in $50$ trials is not $0.009times50$.
  – Jam
  Jul 25 at 10:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If that were the case, the probability it occurs $1$ million times in $1$ million trials would be $9000$, i.e. $900,000%$, which can't be right
  – Jam
  Jul 25 at 10:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If an event had the probability of occurring 0.9% of the time then in 1 million trials it would occur 0.009*1,000,000 times which is 9000 times. What I am trying to say (with your example) that it will occur 9000 times in one million trials. Because the probability of an event occurring is it's probability multiplied by the # of trials. I may not have been very clear with that.
  – Vivek
  Jul 25 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That works for finding the expected number of successes. And I agree that if you perform $n$ trials which are independent and each have a probability of success of $p$ then the expected number of successes is $np$. But the probability of success in all trials would be $P(t_1wedge t_2wedge ldots wedge t_n)=P(t_1)times P(t_2)ldots times P(t_n)=ldots$, where $P(t_i)$ is the probability of success in trial $i$. In your answer you've calculated an expected value, as if it were a probability
  – Jam
  Jul 25 at 13:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Intuitively, you can see that your answer must be flawed since you've got a higher probability when you have a success multiple times. Imagine if your probability of winning the lottery were $0.9%$. Would it not seem strange if your probability of winning the lottery $50$ times consecutively were $4.5%$?
  – Jam
  Jul 25 at 13:47
  
  

  
  
  
  

 | 
show 1 more comment

up vote
-1
down vote

up vote
-1
down vote

For starters 0.0009^50 is an extremely small amount and is so small that most calculators would regard it as 0 (this is because the product of any number less than one and another number less than one is smaller than the original number, for example 0.2*0.2 = 0.04 and 0.04<0.2) . As a result, 2-0.0009^50 is equal to 2.

0.9% is equal to 9/1000. So the probability of an event which occurs once with the probability of 0.9% is 1*9/1000. The probability of an event occurring 50 times with the probability of 0.9% is 50*9/1000 which equals 450/1000 and is simplified to $boxeddfrac9200$ or $boxed4.5 %$.

share|cite|improve this answer

edited Jul 25 at 13:52

answered Jul 25 at 10:00

Vivek

163

For starters 0.0009^50 is an extremely small amount and is so small that most calculators would regard it as 0 (this is because the product of any number less than one and another number less than one is smaller than the original number, for example 0.2*0.2 = 0.04 and 0.04<0.2) . As a result, 2-0.0009^50 is equal to 2.

0.9% is equal to 9/1000. So the probability of an event which occurs once with the probability of 0.9% is 1*9/1000. The probability of an event occurring 50 times with the probability of 0.9% is 50*9/1000 which equals 450/1000 and is simplified to $boxeddfrac9200$ or $boxed4.5 %$.

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edited Jul 25 at 13:52

answered Jul 25 at 10:00

Vivek

163

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 13:52

edited Jul 25 at 13:52

edited Jul 25 at 13:52

answered Jul 25 at 10:00

Vivek

163

answered Jul 25 at 10:00

Vivek

163

answered Jul 25 at 10:00

Vivek

163

163

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  If the probability of an event is $0.009$, then the probability it occurs $50$ times in $50$ trials is not $0.009times50$.
  – Jam
  Jul 25 at 10:09
  

  
  
  
  


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  If that were the case, the probability it occurs $1$ million times in $1$ million trials would be $9000$, i.e. $900,000%$, which can't be right
  – Jam
  Jul 25 at 10:12
  

  
  
  
  


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  If an event had the probability of occurring 0.9% of the time then in 1 million trials it would occur 0.009*1,000,000 times which is 9000 times. What I am trying to say (with your example) that it will occur 9000 times in one million trials. Because the probability of an event occurring is it's probability multiplied by the # of trials. I may not have been very clear with that.
  – Vivek
  Jul 25 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That works for finding the expected number of successes. And I agree that if you perform $n$ trials which are independent and each have a probability of success of $p$ then the expected number of successes is $np$. But the probability of success in all trials would be $P(t_1wedge t_2wedge ldots wedge t_n)=P(t_1)times P(t_2)ldots times P(t_n)=ldots$, where $P(t_i)$ is the probability of success in trial $i$. In your answer you've calculated an expected value, as if it were a probability
  – Jam
  Jul 25 at 13:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Intuitively, you can see that your answer must be flawed since you've got a higher probability when you have a success multiple times. Imagine if your probability of winning the lottery were $0.9%$. Would it not seem strange if your probability of winning the lottery $50$ times consecutively were $4.5%$?
  – Jam
  Jul 25 at 13:47
  
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If the probability of an event is $0.009$, then the probability it occurs $50$ times in $50$ trials is not $0.009times50$.
  – Jam
  Jul 25 at 10:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If that were the case, the probability it occurs $1$ million times in $1$ million trials would be $9000$, i.e. $900,000%$, which can't be right
  – Jam
  Jul 25 at 10:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If an event had the probability of occurring 0.9% of the time then in 1 million trials it would occur 0.009*1,000,000 times which is 9000 times. What I am trying to say (with your example) that it will occur 9000 times in one million trials. Because the probability of an event occurring is it's probability multiplied by the # of trials. I may not have been very clear with that.
  – Vivek
  Jul 25 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That works for finding the expected number of successes. And I agree that if you perform $n$ trials which are independent and each have a probability of success of $p$ then the expected number of successes is $np$. But the probability of success in all trials would be $P(t_1wedge t_2wedge ldots wedge t_n)=P(t_1)times P(t_2)ldots times P(t_n)=ldots$, where $P(t_i)$ is the probability of success in trial $i$. In your answer you've calculated an expected value, as if it were a probability
  – Jam
  Jul 25 at 13:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Intuitively, you can see that your answer must be flawed since you've got a higher probability when you have a success multiple times. Imagine if your probability of winning the lottery were $0.9%$. Would it not seem strange if your probability of winning the lottery $50$ times consecutively were $4.5%$?
  – Jam
  Jul 25 at 13:47
  
  

  
  
  
  

If the probability of an event is $0.009$, then the probability it occurs $50$ times in $50$ trials is not $0.009times50$.
– Jam
Jul 25 at 10:09

If the probability of an event is $0.009$, then the probability it occurs $50$ times in $50$ trials is not $0.009times50$.
– Jam
Jul 25 at 10:09

If that were the case, the probability it occurs $1$ million times in $1$ million trials would be $9000$, i.e. $900,000%$, which can't be right
– Jam
Jul 25 at 10:12

If that were the case, the probability it occurs $1$ million times in $1$ million trials would be $9000$, i.e. $900,000%$, which can't be right
– Jam
Jul 25 at 10:12

If an event had the probability of occurring 0.9% of the time then in 1 million trials it would occur 0.009*1,000,000 times which is 9000 times. What I am trying to say (with your example) that it will occur 9000 times in one million trials. Because the probability of an event occurring is it's probability multiplied by the # of trials. I may not have been very clear with that.
– Vivek
Jul 25 at 13:25

If an event had the probability of occurring 0.9% of the time then in 1 million trials it would occur 0.009*1,000,000 times which is 9000 times. What I am trying to say (with your example) that it will occur 9000 times in one million trials. Because the probability of an event occurring is it's probability multiplied by the # of trials. I may not have been very clear with that.
– Vivek
Jul 25 at 13:25

That works for finding the expected number of successes. And I agree that if you perform $n$ trials which are independent and each have a probability of success of $p$ then the expected number of successes is $np$. But the probability of success in all trials would be $P(t_1wedge t_2wedge ldots wedge t_n)=P(t_1)times P(t_2)ldots times P(t_n)=ldots$, where $P(t_i)$ is the probability of success in trial $i$. In your answer you've calculated an expected value, as if it were a probability
– Jam
Jul 25 at 13:43

That works for finding the expected number of successes. And I agree that if you perform $n$ trials which are independent and each have a probability of success of $p$ then the expected number of successes is $np$. But the probability of success in all trials would be $P(t_1wedge t_2wedge ldots wedge t_n)=P(t_1)times P(t_2)ldots times P(t_n)=ldots$, where $P(t_i)$ is the probability of success in trial $i$. In your answer you've calculated an expected value, as if it were a probability
– Jam
Jul 25 at 13:43

Intuitively, you can see that your answer must be flawed since you've got a higher probability when you have a success multiple times. Imagine if your probability of winning the lottery were $0.9%$. Would it not seem strange if your probability of winning the lottery $50$ times consecutively were $4.5%$?
– Jam
Jul 25 at 13:47

Intuitively, you can see that your answer must be flawed since you've got a higher probability when you have a success multiple times. Imagine if your probability of winning the lottery were $0.9%$. Would it not seem strange if your probability of winning the lottery $50$ times consecutively were $4.5%$?
– Jam
Jul 25 at 13:47

 | 
show 1 more comment

 

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