Probability of <3 Errors in 10-char Msg

AI Thread Summary
The discussion focuses on calculating the probability of encountering less than 3 errors in a 10-character message, where each character has a 1/10 chance of being erroneous. The binomial formula is applied correctly, using n = 10, p = 1/10, and summing the probabilities for k = 0, 1, and 2 errors. Participants confirm that the approach is valid, emphasizing the importance of accurate arithmetic in the calculations. The conversation reassures that the method used is appropriate for solving the problem. Overall, the probability of less than 3 errors is derived using the binomial distribution.
magnifik
Messages
350
Reaction score
0
In sending 10 characters, a character error occurs independently with probability 1/10. What is the probability that in a 10-character message, less than 3 errors occur?

I am using the binomial formula (n choose k)pk(1-p)n-k where n = 10, p = 1/10, and k is the number of errors. Since the problem statement says less than 3 errors occur, I adding up the values for k = 0, 1, 2

(10 choose 0)(1/10)0(1-1/10)10 + (10 choose 1)(1/10)1(1-1/10)9 + (10 choose 2)(1/10)2(1-1/10)8, but i am wondering if I am doing this correctly? should i be adding or multiplying?
 
Physics news on Phys.org
magnifik said:
In sending 10 characters, a character error occurs independently with probability 1/10. What is the probability that in a 10-character message, less than 3 errors occur?

I am using the binomial formula (n choose k)pk(1-p)n-k where n = 10, p = 1/10, and k is the number of errors. Since the problem statement says less than 3 errors occur, I adding up the values for k = 0, 1, 2

(10 choose 0)(1/10)0(1-1/10)10 + (10 choose 1)(1/10)1(1-1/10)9 + (10 choose 2)(1/10)2(1-1/10)8, but i am wondering if I am doing this correctly?

You have written everything correctly, so if you have not made any arithmetical errors your answer should be correct.

RGV
 
thanks
 
Last edited:
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top