1. ## a question about discrete RV

hello,
I have a question about discrete RV.
Five fair coins are tossed. The coins that fall on "Head" are removed and the rest of them are tossed again. Then we removed again the coins that fall on "Head" and so forth until all the coins are removed.
What is the probability that after the fifth round not all the coins were removed?
Thanks a lot,
Avshi

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studentttt (08-07-2014)

3. ## Re: a question about discrete RV

Where are you having troubles with this question?

4. ## Re: a question about discrete RV

I tried to solve it but I couldn't, so I'm also interested in the solution to learn something. I started with

X(1) = number of coins remained after the first run
X(2) = number of coins remained after the second run
....

P(X(1) > 0) = P(at least one coin tossed tail)
= 1 - 1/32
P(X(2) > 0) = P(X(1) > 0) * P(at least on coin in the second run is tail)

But I don't know how to get P(at least on coin in the second run is tail). Or my attempt is wrong in the very beginning. Please advise.

5. ## Re: a question about discrete RV

Try to think in the following way:

Let be the number of tosses the -th coin have before it is removed.

What is its distribution? (actually it is not a must to know this answer; it just act as a side question to consolidate the concept)

How does it relate to your original question?

6. ## Re: a question about discrete RV

Let me try.... X(i) is a binomial distribution, Bin(5,0.5).

Let X(i) be the number of heads after the experiment.

P(after fifth not all coins are removed) = P(first coin all tail) + ... + P(fifth coin all tail)

= P(X(1)=0) + ... + P(X(5)=0)

where X(i) = Bin(5,0.5)

7. ## Re: a question about discrete RV

Sorry it is incorrect. Actually is the number of tosses to obtain the first head.

8. ## Re: a question about discrete RV

X(i) = number of tosses to obtain the first head

Probability after the fifth round that not all coins are head = P(X(1)>5) + ... + P(X(5)>5) = 5 * P(X(1)>5) because the tosses are independent

P(X(1)>5) = 1 - P(X(1)<=5) = P(first toss to get a head) + .... + P(fifth toss to get a head)

= (1/2) + (1/2*1/2) + ... + (1/2*1/2*1/2*1/2*1/2) = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 0.9875

Therefore, the answer would be 5 * (1-0.9875).

This looks like a Maclaurin series to me.

9. ## Re: a question about discrete RV

Closer but not exact. The I defined is corresponding to a specific coin. They are independent as you said, but the events are not mutually exclusive. Therefore, you cannot break the probability of union into sum like that. Instead you should try to work on the intersection which you can use the independent property.

10. ## Re: a question about discrete RV

Originally Posted by BGM
Try to think in the following way:

Let be the number of tosses the -th coin have before it is removed.

What is its distribution? (actually it is not a must to know this answer; it just act as a side question to consolidate the concept)

How does it relate to your original question?
Thank you BGM.
if I understand you correctly each xi has a geometric distribution with p=0.5.
the probability that after 5 runs all the coins will remove is 1- p(x1<=5 and x2<=5 and ... x5<=5)=1-p(x1<=5)*p(x2<=5)*...*P(x5<=5)=1-[1-(0.5)^5]^5=0.15
is it correct?

11. ## Re: a question about discrete RV

Originally Posted by BGM
Try to think in the following way:

Let be the number of tosses the -th coin have before it is removed.

What is its distribution? (actually it is not a must to know this answer; it just act as a side question to consolidate the concept)

How does it relate to your original question?
Thank you BGM.
if I understand you correctly each xi has a geometric distribution with p=0.5.
the probability that after 5 runs not all the coins will remove is 1- p(x1<=5 and x2<=5 and ... x5<=5)=1-p(x1<=5)*p(x2<=5)*...*P(x5<=5)=1-[1-(0.5)^5]^5=0.15
is it correct?

12. ## Re: a question about discrete RV

Have not check the numeric answer but the steps look good to me.

13. ## Re: a question about discrete RV

Thanks
Avshi

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