A fair coin is tossed 10 times. What is the probability that

I agree this is not a straight forward binomial distribution question. I think the formula needs to be modified for a question like this, which I tried to do. I am sure there is a mathematical answer to this using some form of a maniuplated binomial distribution equation. Lets discuss this together.

The question is asking you the number of places you can place a success given (0 heads given 10 tails) + (1 heads given 9 tails) + (2 heads 8 tails) + (3 heads 7 tails) + (4 heads 6 tails) + (5 heads 5 tails) in 10 trials of a binomial event, whereas a success is defined as a heads in any spot that is not next to another heads...

Meaning, for (1 head given 9 tails) you have _T_T_T_T_T_T_T_T_T_ 10 spots. You must take the 10 spots and choose 1 of them. How many different ways can you choose that 1 spot? There are 10c1 ways. Or 10!/(10-1)!*1!.]

As you see in the part you quoted of me, I said the "probability of getting 0,1,2,3,4, or 5 heads (successes)" Therefore, the wording is correct, as I am eliminating the failures by not counting spots for them.

Also, if you do correctly use the equation I provided (my explanation tells you that in a case like this the first part of the equation is the SUM of the first part for each success), you get the right answer.

Here is the math:

First half of the equation:

0 heads, 10 tails= TTTTTTTTTT = 0c0
1 Heads, 9 tails = _T_T_T_T_T_T_T_T_T_ = 10c1
2 heads, 8 tails = _T_T_T_T_T_T_T_T_ = 9c2
3 heads, 7 tails = _T_T_T_T_T_T_T_ = 8c3
4 heads, 6 tails = _T_T_T_T_T_T_ = 7c4
5 heads, 5 tails = _T_T_T_T_T_ = 6c5

0c0 + 10c1 + 9c2 + 8c3 + 7c4 + 6c5 =
1 + 10 + 36 + 56 + 35 + 6 =
144

second part of the equation

The second part of the formula, r becomes 144, and N is 10. Why is r 144? Because as from above there are 144 chances of a success.

(1/2)144 * (1/2)^(10-144) =
1/2^144 * 1/2^-134 =
1/2^144 * 2^134 =
2^-10 =
1/2^10 =
1/1024


Put it together


144 * 1/1024 =
144/1024 = answer

144/1024 = probability of getting 0,1,2,3,4, or 5 heads in a successful manor as defined by the question.
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I believe this is right. However I welcome all discussion.

My goal is to provide good and clear explanations for forum members. Sometimes somebody has already given a great explanation, so I try and find an alternative explanation to help further people's understandings.

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Benjiboo

Yes, that's the solution KASSALMD gave in an earlier post (though, as I later pointed out, he omitted the cases of 0 Heads and 1 Head), and it is correct. There's really no need to use a binomial probability formula here - by the definition of probability, we have 1 + 10C1 + 9C2 + 8C3 + 7C4 + 6C5 ways of getting the result we want, and 2^10 possible outcomes in total, so dividing the first number by the second gives our answer. There would only be an advantage to using binomial probability if the probability of getting Heads was different from 1/2.

what if in case 1, all the tosses are tails as well as in case 2. therefore a possibility of all the tosses are tails is counted twice. my point is each possibility must only be counted once not twice. I am not sure though. pls correct me if i am wrong.

Yeah I got the same answer, D specifically 144/1024
But I did it as 1-probability(getting 2 Heads in a row)
I think it was more complicated...but still used fibonacci

2^8+(2^7)*1+(2^6)*2+(2^5)*3+(2^4)*5+(2^3)*8+(2^2)*13+(2^1)*21+(2^0)*34
= 880

Notice the 1, 2, 3, 5, 8, 13, 21, 34 sequence

1-880/1024 = 144/1024
WOw, spent the whole night on this one.

This is great as usual from you

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