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#1 2013-02-17 12:34:35

White_Owl
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Registered: 2010-03-03
Posts: 99

Probabilities: pmf and cdf

I am reading and rereading the textbook and I still do not understand what exactly is the pmf and cdf?

Here is a problem from the last homework:
Suppose we have three couples and two individuals - five independent groups in total (eight people). Three couples are marked #1, #2, and #3, groups of one person are marked #4 and #5. Each of these five groups can be late for a meeting with a probability 40%. All groups are independent from each other.
Let X be a number of people who arrived late.
Determine pmf and cdf for X.

Well, first I tried to formalize X:


Since there are two possible combinations for 2, 4, and 6 people, we need to take a union of probabilities for these combinations and final P(X) become this:

And from here I need to find pmf
All examples in the textbook are just doing direct substitution p(x)=P(X=x) but if I summarize my P(X) I receive 1.4575616. And sum of p(x) for all x is supposed to be equal 1.0...

So, did I make a mistake somewhere or what am I missing?

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#2 2013-02-17 13:20:12

anonimnystefy
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From: The Foundation
Registered: 2011-05-23
Posts: 15,390

Re: Probabilities: pmf and cdf

Those probabilities are not correct. The problem is that if you want exactly one person to be late, you only get the probability that one of the two is late, but not that everybody else isn't.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#3 2013-02-18 11:15:50

White_Owl
Member
Registered: 2010-03-03
Posts: 99

Re: Probabilities: pmf and cdf

I am not sure I understand. If we say that for the "exactly one group is late" we need to calculate: intersection of "one group is late" and "four groups are not late", and any of the five groups can be late. So the formula for the one group become:


But once I recalculate whole table like this - the sum become 1.98.

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#4 2013-02-18 11:34:50

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,390

Re: Probabilities: pmf and cdf

Try using the exact values. And, also, it should be a 2 there, not a 5.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#5 2013-02-18 15:22:00

White_Owl
Member
Registered: 2010-03-03
Posts: 99

Re: Probabilities: pmf and cdf

Ok... Lets do it this way:
P(X)= Combination * (0.4^Total_Groups * 0.6^(5-Total_Groups))

And if I collapse rows with same number of people as a union of the two probabilities I receive:

Still no 1.0 in the total. What is wrong now?

Last edited by White_Owl (2013-02-18 15:22:54)

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#6 2013-02-18 20:30:23

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,390

Re: Probabilities: pmf and cdf

Hi

I am getting 1 for the total in the first table.

It seems to me you forgot to add the case when X=0.

Last edited by anonimnystefy (2013-02-18 20:35:18)


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#7 2013-02-19 02:46:16

White_Owl
Member
Registered: 2010-03-03
Posts: 99

Re: Probabilities: pmf and cdf

oh.... yes.
Then I guess the problem is solved.
Thank you.

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#8 2013-02-19 06:43:15

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,390

Re: Probabilities: pmf and cdf

You're welcome.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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