# Probability question

• Jan 29th 2011, 06:17 PM
Mykmahon
Probability question
Hello math forum.
I have a question.
What are the odds of a number between 1 and 152 being selected 6 Times in 15 chances?
• Jan 29th 2011, 06:25 PM
snowdrifter
Quote:

Originally Posted by Mykmahon
Hello math forum.
I have a question.
What are the odds of a number between 1 and 152 being selected 6 Times in 15 chances?

Assuming you mean exactly 6 times the answer would be...
The odds of being chosen 6 times multiplied by the odds of NOT being chosen 9 times.
• Jan 29th 2011, 06:30 PM
Plato
Quote:

Originally Posted by Mykmahon
Hello math forum.
I have a question.
What are the odds of a number between 1 and 152 being selected 6 Times in 15 chances?

Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service.
PLease either post some of your own work on this problem or explain what you do not understand about the question.
• Jan 29th 2011, 06:38 PM
Mykmahon
I do not know how to figure it out. I need a formula or something to figure it out ifaonebody can't answer it for me. Thank you for any help
• Jan 29th 2011, 07:03 PM
Mykmahon
I dint know how to figure out the oddsof being pulled 6 times or the odds if not being pulled 9 times. Itis exactly 6 times I am trying to figure out.
• Jan 29th 2011, 09:09 PM
mr fantastic
Quote:

Originally Posted by Mykmahon
Hello math forum.
I have a question.
What are the odds of a number between 1 and 152 being selected 6 Times in 15 chances?

Do you mean "What are the odds of the same number between 1 and 152 being selected 6 Times in 15 chances?"
• Jan 29th 2011, 10:24 PM
Mykmahon
Quote:

Originally Posted by mr fantastic
Do you mean "What are the odds of the same number between 1 and 152 being selected 6 Times in 15 chances?"

Yes. The same number
• Jan 30th 2011, 05:07 AM
Plato
The reason we ask you to show some work is to get some idea what the difficulty you have. Your question can be read at least two ways.
1) Given a number from 1 to 152, what is the probability that number would be randomly and independently selected exactly six times in 15 chances?
OR
2) A number is selected from 1 to 152, what is the probability that same number would be randomly and independently selected exactly 5 more times in 14 chances?

The answers are similar but not the same.

1) $\dbinom{15}{6}\left(\dfrac{1}{152}\right)^{6}\left (\dfrac{151}{152}\right)^{9}$

2) $\dbinom{14}{5}\left(\dfrac{1}{152}\right)^{5}\left (\dfrac{151}{152}\right)^{9}$
• Jan 30th 2011, 02:15 PM
Mykmahon
Quote:

Originally Posted by Plato
The reason we ask you to show some work is to get some idea what the difficulty you have. Your question can be read at least two ways.
1) Given a number from 1 to 152, what is the probability that number would be randomly and independently selected exactly six times in 15 chances?
OR
2) A number is selected from 1 to 152, what is the probability that same number would be randomly and independently selected exactly 5 more times in 14 chances?

The answers are similar but not the same.

1) $\dbinom{15}{6}\left(\dfrac{1}{152}\right)^{6}\left (\dfrac{151}{152}\right)^{9}$

2) $\dbinom{14}{5}\left(\dfrac{1}{152}\right)^{5}\left (\dfrac{151}{152}\right)^{9}$

I appreciate the help with the formula for your number one. That is the answer I want but I can't do the math on my calculator. I don't know how to do the power of 6 or 9 parts. I would really appreciate the answer being posted or how to do this problem on a scientific calculator. Thanks for any help
• Jan 30th 2011, 02:25 PM
Plato
1) $\dbinom{15}{6}\left(\dfrac{1}{152}\right)^{6}\left (\dfrac{151}{152}\right)^{9}=0.000000000382422$
• Jan 30th 2011, 03:14 PM
Mykmahon
Quote:

Originally Posted by Plato
1) $\dbinom{15}{6}\left(\dfrac{1}{152}\right)^{6}\left (\dfrac{151}{152}\right)^{9}=0.000000000382422$

How would that answer be expressed in ratio form?? As in the odds of winning something. Ex-- 3 to 1

3.8244 billion to 1. ???
• Jan 30th 2011, 04:57 PM
Mykmahon
Quote:

Originally Posted by Plato
1) $\dbinom{15}{6}\left(\dfrac{1}{152}\right)^{6}\left (\dfrac{151}{152}\right)^{9}=0.000000000382422$

I got two of the three numbers I need however, I do not know how to read or factor in the number answer for 1/152 to the sixth power. My calculator shows it as this

8.1084617495525e-14

What do I do with this number? I can't multiply that by the other two numbers in the equation which are

15/6 = 2.5
151/152 to the ninth power = .942323960625871

How do I compute these three numbers using scientific calculator??
• Jan 31st 2011, 01:06 AM
mr fantastic
Quote:

Originally Posted by Mykmahon
I appreciate the help with the formula for your number one. That is the answer I want but I can't do the math on my calculator. I don't know how to do the power of 6 or 9 parts. I would really appreciate the answer being posted or how to do this problem on a scientific calculator. Thanks for any help

How to use your calculator is best discussed with your teacher because I cannot see what trouble there could be in keying the required numbers in.