A stick of length 1 is divided randomly into 3 parts.
What is the probability that a triangle can be made with those three parts?
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2 Answers
62Woow u have posted like what 5-6 questions..
I will try to answer each of them one by one..
This will need u to draw a graph..
Of :
0<x<0.5
0<y<0.5
0<|x-y|<0.5
Total area will be inside 0<x<1 and 0<y<1
I think this is enough. I know this solution is very unconventional.. But i hope u get what i try to mean :)
24A triangle can be made, if and only if,it satisfies triangle inequality
=>a < b + c
b < c + a
c < a + b
Also, a +b+ c = 1
=> a < 1/2, b < 1/2,c < 1/2
=>A triangle can be formed, if all three sides < 1/2 and sum is 1.
Now, let's find the probability that one of a, b, c ≥1/2.
Note that to divide stick randomly into 3 parts, we need to choose two numbers P and Q, both are between 0 & 1 and P
Now, a ≥ 1/2, iff both the numbers, P & Q≥1/2.
=> probability of a≥ 1/2 = (1/2) * (1/2) = 1/4
Similarly, c ≥ 1/2, iff P & Q≤ 1/2.
=> probability of c≥ 1/2 =(1/2) * (1/2) = 1/4
Also, probability of b ≥ 1/2 is = (1/2) * (1/2) = 1/4
The probability that a triangle can not be made
= (1/4) + (1/4) + (1/4)
= (3/4)
Thus, the probability that a triangle can be made
= 1 - (3/4)
= (1/4)
=> the probability that a triangle can be made by randomly dividing a stick of length 1 into 3 parts is 25%