Question 1188499
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Let X represent the difference between the number of heads and the number of tails 
when a coin is tossed 41 times. Then P(X=11)=
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It is very nice,  slightly tangled probability problem.

I solved it  (a  TWIN  version)  once at this forum;  see my lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Challenging-problems-on-Binomial-distribution-probability.lesson>Challenging problems on Binomial distribution probability</A> 

(Problem 1) &nbsp;in this site.  &nbsp;So, &nbsp;it is the second time I meet this problem at the forum . . . 



<pre>
Let H be the number of heads and T be the number of tails.

Then we have these two equations

    H + T = 41

    H - T = 11


We solve this system elementary, using elimination method, and we get H = {{{(41+11)/2}}} = 26;  T = 15.


Now the problem is: find the probability to get 26 heads tossing a coin 41 times.


The answer is  P = {{{C[41]^26*0.5^25*0.5^15}}} = {{{63432274896*0.5^41}}} = 0.028845659 = 0.02885 (rounded).  
</pre>

Solved.