Probability of <3 Errors in 10-char Msg

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SUMMARY

The probability of encountering less than 3 errors in a 10-character message, where each character has an independent error probability of 1/10, can be calculated using the binomial formula. The correct approach involves summing the probabilities for k = 0, 1, and 2 errors using the formula (n choose k)pk(1-p)n-k. Specifically, the calculations involve (10 choose 0)(1/10)0(9/10)10 + (10 choose 1)(1/10)1(9/10)9 + (10 choose 2)(1/10)2(9/10)8. If executed without arithmetic errors, this method yields the correct probability.

PREREQUISITES
  • Understanding of binomial probability distribution
  • Familiarity with the binomial formula (n choose k)pk(1-p)n-k
  • Basic knowledge of combinatorics
  • Ability to perform arithmetic calculations accurately
NEXT STEPS
  • Study the binomial distribution in depth
  • Learn about the implications of independent events in probability
  • Practice solving problems using the binomial formula
  • Explore applications of probability theory in error detection and correction
USEFUL FOR

Students, educators, and professionals in statistics, data science, and computer science who are interested in understanding error probabilities in data transmission.

magnifik
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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?
 
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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:

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