• I am having a problem with a probability question. It goes as follows: In a lottery problem, 4 numbers are always drawn from a possible space of 9 numbers
Message 1 of 7 , Mar 15 12:16 PM
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I am having a problem with a probability question. It goes as follows:

In a lottery problem, 4 numbers are always drawn from a possible space
of 9 numbers between 0 and 9. What is the probability that if a person
pre-selects 4 numbers that

a) 4 of the 4 will be in correct order of selection

b) 3 of 4 will be in correct order of selection

c) 2 of 4 will be in correct order of selection

Thank you
• ... Samuel, I would like to help, but I am not sure I understand the questions. Could you be more explicit as to what the experiment is? Regards. Noel.
Message 2 of 7 , Mar 16 12:18 PM
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> I am having a problem with a probability question. It goes as follows:
>
> In a lottery problem, 4 numbers are always drawn from a possible space
> of 9 numbers between 0 and 9. What is the probability that if a person
> pre-selects 4 numbers that
>
> a) 4 of the 4 will be in correct order of selection
>
> b) 3 of 4 will be in correct order of selection
>
> c) 2 of 4 will be in correct order of selection
>
> Thank you

Samuel,
I would like to help, but I am not sure I understand the questions.
Could you be more explicit as to what the experiment is?

Regards. Noel.
• For a series of dependent trials the probability of success on any trial is (k+1)/(k+2) where k is equal to the number of successes on the previous two trials.
Message 3 of 7 , Feb 8, 2004
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For a series of dependent trials the probability of success on any
trial is (k+1)/(k+2) where k is equal to the number of successes on
the previous two trials. Compute
lim (n -> infinity) P{success on the nth trial}
• Sometimes for limits, as in the following case, you have a shortcut to finding P(n), provided you assume the limit required exists. Say the limit exists, and
Message 4 of 7 , Feb 9, 2004
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Sometimes for limits, as in the following case, you have a shortcut to
finding P(n), provided you assume the limit required exists.

Say the limit exists, and is q. Then, the expected number of successes on
the "previous two trials" for large n must be 2q. Hence,

q = (2q+1)/(2q+2)
=> q = 1/Sqrt(2)

Hope that helps.

vikasgupta2k <vikasgupta2k@...>
09/02/2004 08:21 AM
probability@yahoogroups.com

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Subject
[probability] Probability question

For a series of dependent trials the probability of success on any
trial is (k+1)/(k+2) where k is equal to the number of successes on
the previous two trials. Compute
lim (n -> infinity) P{success on the nth trial}

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• Can someone help me with this? A bag contains 3 yellow marbles, 4 red marbles and 1 green marble. One marble is taken out from the bag and its colour recorded.
Message 5 of 7 , Sep 16, 2005
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Can someone help me with this?

A bag contains 3 yellow marbles, 4 red marbles and 1 green marble.
One
marble is taken out from the bag and its colour recorded. It is then
returned to the bag before the next marble is taken out. This process
is repeated 3 times. What is the probability that the three marbles
drawn
(a) are of the same colour?
(b) are all different in colour?
• ... a: P (the same color)= P(3 yellows)+P(3 reds)+P(3 green)= (3/8)^3+(4/8)^3+0. b)6*(3/8)*(4/8)*(1/8)
Message 6 of 7 , Oct 4, 2005
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--- In probability@yahoogroups.com, "josephine_jalleh"
<josephine_jalleh@h...> wrote:
> Can someone help me with this?
>
> A bag contains 3 yellow marbles, 4 red marbles and 1 green marble.
> One
> marble is taken out from the bag and its colour recorded. It is then
> returned to the bag before the next marble is taken out. This process
> is repeated 3 times. What is the probability that the three marbles
> drawn
> (a) are of the same colour?
> (b) are all different in colour?

a: P (the same color)= P(3 yellows)+P(3 reds)+P(3 green)=

(3/8)^3+(4/8)^3+0.

b)6*(3/8)*(4/8)*(1/8)
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