Probability of getting an ace in case of a loaded die

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

12)a)
a) Since both events are independent : P( 33) = 9 / 441

b) Let’s have

4: throwing 4

Not 4 : not throwing 4

P ( 4 and not throwing 4 ) = P( 4) P ( not throwing 4) = (17*4)/441c) Let’s consider the sample event S for the above problem:

S : {(6,4),(4,6),(5,6),(6,5),(5,5),(6,6)}

Probability of getting 5 and 5 under given condition is : P(55/S)

Using Bayes’ theorem,

P(55/S) = P(55 and S)/ P(S)As S contains (5,5) as a sample point, P(55 and S) = P(55)

Is it always true that P (A and B) = P (A),when A is a subset of B ?P(55/S) = $\frac { \frac {5 *5 } {21*21 } } {(2*6*4 +2*5*6 +25+36)/441 }$

= $\frac { 25 } { 169 }$d) I think what the question says is:

I throw the dice n times, what is the probability that I will get an ace once?
The dice throwing n times and getting an ace once is equivalent to that I have n boxes and I have to put one ace in any of these n boxes. The probability of getting an ace is 1/21 in one throw. So, the probability of getting one ace in n throws is n/21.Now, $\frac { n } {21 } \geq ½ \\ n \geq 11$
e) S = {(2,5),(2,6), (4,5),(4,6), (6,5),(6,6)}

P(S) = 132/441

Is this correct ?

 

[.Scott]

A) Simplfiy the 9/441
B) Good
C) Good
D) The probability of not getting an ace in 1 throw is 20/21. In N throws, the probability of no ace is (20/21)^N.
E) Simplify 132/441

 

[Pushoam]

D) The probability of not getting an ace in 1 throw is 20/21. In N throws, the probability of no ace is (20/21)^N.

The probability of getting not " no ace" i.e. at least one ace in N throws is:
1- ${\frac{20}{21}}^N$.
Now, 1- ${\frac{20}{21}}^N \geq 0.5 \\ N = 15$
But, this will tell me probabilty of getting at least one ace, not exacctly one ace.

 

[BvU]

Hey, didn't my reply come through ? Weird. Your logic in d) is the same as claiming you have 100% chance of throwing a six in six throws with an unloaded die.

In the problem wording 'an ace' means 'not zero aces'

 

[Ray Vickson]

The probability of getting not " no ace" i.e. at least one ace in N throws is:
1- ${\frac{20}{21}}^N$.
Now, 1- ${\frac{20}{21}}^N \geq 0.5 \\ N = 15$
But, this will tell me probabilty of getting at least one ace, not exacctly one ace.

Let $N$ be the number of tosses you need to make to get your first ace. That follows a geometric distribution with "success" probability $p = 1/21$:
$$P(N = n) = p (1-p)^{n-1}, n = 1,2, \ldots,$$
and
$$P(N > n) = (1-p)^n, n=0,1,2, \ldots .$$
In other words, probability of needing more than $n$ tosses to your first ace is $(20/21)^n$, because the first $n$ tosses must all end in a non-ace.

Be careful If you Google "geometric distribution", because some sources (such as http://mathworld.wolfram.com/GeometricDistribution.html ) give the "alternative" form, which is the number of tosses before your first ace. That is essentially $N_b = N-1$, and has distribution $P(N_b = n) = p (1-p)^n, n=0,1,2, \ldots$

 

[haruspex]

Let $N$ be the number of tosses you need to make to get your first ace. That follows a geometric distribution with "success" probability $p = 1/21$:
$$P(N = n) = p (1-p)^{n-1}, n = 1,2, \ldots,$$
and
$$P(N > n) = (1-p)^n, n=0,1,2, \ldots .$$
In other words, probability of needing more than $n$ tosses to your first ace is $(20/21)^n$, because the first $n$ tosses must all end in a non-ace.

Be careful If you Google "geometric distribution", because some sources (such as http://mathworld.wolfram.com/GeometricDistribution.html ) give the "alternative" form, which is the number of tosses before your first ace. That is essentially $N_b = N-1$, and has distribution $P(N_b = n) = p (1-p)^n, n=0,1,2, \ldots$

Are you implying that @Pushoam 's response is incorrect somewhere, either in logic or result? It looks fine to me, and a bit simpler than thinking terms of throws until the first Ace.
And it produces the same answer, yes?

 

[Ray Vickson]

Are you implying that @Pushoam 's response is incorrect somewhere, either in logic or result? It looks fine to me, and a bit simpler than thinking terms of throws until the first Ace.
And it produces the same answer, yes?

I'm not implying anything; I am just responding to his/her statement "But, this will tell me probabilty of getting at least one ace, not exactly one ace" in Post #3. The probability of getting at least one ace is exactly what is needed in this problem; the probability of getting exactly 1 ace in $n$ tosses will not answer the original question.

 

[haruspex]

I'm not implying anything; I am just responding to his/her statement "But, this will tell me probabilty of getting at least one ace, not exactly one ace" in Post #3. The probability of getting at least one ace is exactly what is needed in this problem; the probability of getting exactly 1 ace in $n$ tosses will not answer the original question.

Ok, thanks for clarifying.

 

[Pushoam]

Thanks for clarifying.

I am not getting the quote, reply, post a new thread button. What should I do now?
Please help me.
Pushoam

 

[haruspex]

Thanks for clarifying.

I am not getting the quote, reply, post a new thread button. What should I do now?
Please help me.
Pushoam

Have you tried reloading the page?

 

[Divyazen]

Have you tried reloading the page?

Yes, l did. It seems as if I have got banned.
How can I get to know whether it is temporary or permanent?
And why did it happen?
I want to use my account.
For how long will it remain banned?
Do they not inform before baning?
To whom should l request to get the permission?

 

[berkeman]

Yes, l did. It seems as if I have got banned.
How can I get to know whether it is temporary or permanent?
And why did it happen?
I want to use my account.
For how long will it remain banned?
Do they not inform before baning?
To whom should l request to get the permission?

I explained it to you in my PM a couple yours ago today. Do not create alternate accounts to try to get around the 10-day temporary ban. You have accumulated too many infraction points for many problem posts, and that lead to the temporary ban. Thread is closed.