- #1
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?
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?