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For one toss of a certain coin, the probability that the out
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23 Jul 2008, 02:07
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For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times? A. (0.6)^5 B. 2(0.6)^4 C. 3(0.6)^4 D. 4(0.6)^4(0.4) + (0.6)^5 E. 5(0.6)^4(0.4) + (0.6)^5 OPEN DISCUSSION OF THIS QUESTION IS HERE: foronetossofacertaincointheprobabilitythattheout135559.html
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Re: For one toss of a certain coin, the probability that the out
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23 Jul 2008, 02:12
tarek99 wrote: For one toss of a certain coin, the probability that the outcome is heads is \(0.6\). If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
a) \(0.6^5\)
b) \(2(.6)^4\)
c) \(3(0.6)^4 (0.4)\)
d) \(4(0.6) (0.4) + (0.6)^5\)
e) \(5(0.6)^4 (0.4) + (0.6)^5\)
please explain 4H1T + 5H (HHHHT + HHHTH + HHTHH + HTHHH + THHHH) + (HHHHH) 5 * 0.6^4 * 0.4 + 0.6^5 Option E



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Re: For one toss of a certain coin, the probability that the out
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23 Jul 2008, 03:46
E for me as well, using binomial theorem



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Re: For one toss of a certain coin, the probability that the out
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23 Jul 2008, 16:45
Hello, can you tell me why you are multiplying it by 5? 5 * 0.6^4 * 0.4 + 0.6^5 Thanks. durgesh79 wrote: tarek99 wrote: For one toss of a certain coin, the probability that the outcome is heads is \(0.6\). If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?
a) \(0.6^5\)
b) \(2(.6)^4\)
c) \(3(0.6)^4 (0.4)\)
d) \(4(0.6) (0.4) + (0.6)^5\)
e) \(5(0.6)^4 (0.4) + (0.6)^5\)
please explain 4H1T + 5H (HHHHT + HHHTH + HHTHH + HTHHH + THHHH) + (HHHHH) 5 * 0.6^4 * 0.4 + 0.6^5 Option E



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Re: For one toss of a certain coin, the probability that the out
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23 Jul 2008, 17:29
0.6^5 + 5C4 * (0.6^4) * (0.4)
= 0.6^5 + 5 * (0.6^4) * 0.4
E



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Re: For one toss of a certain coin, the probability that the out
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23 Jul 2008, 18:04
Choices with 4 heads & 1 tail: 5C4 = 5 = 5*0.4*0.6\(4\) Choice with all heads: 0.6\(5\) 0.6\(5\) + 5*0.4*0.6\(4\) E



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Re: For one toss of a certain coin, the probability that the out
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23 Jul 2008, 19:37
so you have to multiply by the #different combinations even though order doesn't matter??



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Re: For one toss of a certain coin, the probability that the out
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24 Jul 2008, 03:06
pmenon wrote: E for me as well, using binomial theorem could you expand on that please? Thanx



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Re: For one toss of a certain coin, the probability that the out
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24 Jul 2008, 04:46
Thank you for your explanations. Makes sense now.



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Re: For one toss of a certain coin, the probability that the out
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24 Jul 2008, 15:10
Explanation of general formula for those who asked:
Imagine that we have an experiment when the coin is tossed n times. Probability of heads in each toss is p, probability of tails is q (clearly q+p=1).
Then, the probability that from n tosses, there will be exactly k heads can be calculated as
P(k heads from n tosses) = C(k,n)*p^k*q^(nk)
(can be proved from the reasoning that 1^n=(p+q)^n=p^n+C(1,n)*p^(n1)*q + ... +C(n1,n)*p*q^(n1)+q^n  the binomial coefficients are there, thus the name)
In this problem, we need to calculate P(4 from 5)+ P(5 from 5) = [use the formula] = 5*(0.6)^4*0.4+1*(0.6)^5, which is E.
You might check wiki for 'binomial distribution' or 'binomial theorem' as well.



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Re: For one toss of a certain coin, the probability that the out
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24 Jul 2008, 17:52
Alternatively,
Let X = number of times it turns up Heads (w.p. 0.6)
X~Binomial(5,0.6)
Pr (X>=4) = Pr(X=4) + Pr(X=5) = (5C4)(0.6)^4 0.4 + (5C5)0.6^5
which is E.



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Re: For one toss of a certain coin, the probability that the out
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24 Jul 2008, 18:51
greenoak wrote: Explanation of general formula for those who asked:
Imagine that we have an experiment when the coin is tossed n times. Probability of heads in each toss is p, probability of tails is q (clearly q+p=1).
Then, the probability that from n tosses, there will be exactly k heads can be calculated as
P(k heads from n tosses) = C(k,n)*p^k*q^(nk)
(can be proved from the reasoning that 1^n=(p+q)^n=p^n+C(1,n)*p^(n1)*q + ... +C(n1,n)*p*q^(n1)+q^n  the binomial coefficients are there, thus the name)
In this problem, we need to calculate P(4 from 5)+ P(5 from 5) = [use the formula] = 5*(0.6)^4*0.4+1*(0.6)^5, which is E.
You might check wiki for 'binomial distribution' or 'binomial theorem' as well. Thanks. It looks complicated. I will stick to my approach



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Re: For one toss of a certain coin, the probability that the out
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Re: For one toss of a certain coin, the probability that the out
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27 Nov 2018, 10:43
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