Question 1145144
.


From the first glance,  it is clear that the given info/input is  NOT  SUFFICIENT  to solve the problem - so,  appropriate 
auxiliary assumptions should be made.  


Moreover, &nbsp;making appropriate and likelihood assumptions is the &nbsp;<U>PART of the solution</U>.



<pre>
Having nothing else given, an appropriate hypothesis is to assume that we have continuous random variable 


    X = "the time of a Brand LT5 battery service / (or "life")" 


with random values that are uniformly distributed with the minimum value of 0 and the maximum value of 200 hours.


Then the mean average  is exactly 100, as the problem states, and  the condition becomes "consistent with the given part" and "self-closed".


Then the probability  P(X < 100) = {{{1/2}}}, as well as  P(x > 100) = {{{1/2}}}.


Therefore, the probability that at least two of the three batteries will last longer than 100 hours is  {{{C[3]^2*(1/2)^3}}} + {{{C[3]^3*(1/2)^3}}} = {{{3/8}}} + {{{1/8}}} = {{{4/8}}} = {{{1/2}}}.
</pre>

Solved.


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To see many other similar solved problems, look into the lesson

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

in this site.


In particular, this problem is VERY SIMILAR to the solved Problem 4 of that lesson.