studentttt (08-07-2014)
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
studentttt (08-07-2014)
Where are you having troubles with this question?
Stop cowardice, ban guns!
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 - P(all heads)
= 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.
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)
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.
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.
Last edited by avsha43; 08-08-2014 at 07:25 AM.
Have not check the numeric answer but the steps look good to me.
Thanks
your explanation was very useful.
Avshi
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