Question : The probability of error in the transmission of a bit over a communication channel is $\displaystyle p = 10^{-4}$. What is the probability of more than three errors in transmitting a block of 1,000 bits?
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It's a binomial distribution. $\displaystyle _{1000}C_x (10^{-4})^x (1 - 10^{-4})^{1000-x}$ So P(X>3) = 1 - P(X <=3) where P(X <= 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
Originally Posted by statmajor It's a binomial distribution. $\displaystyle _{1000}C_x (10^{-4})^x (1 - 10^{-4})^{1000-x}$ So P(X>3) = 1 - P(X <=3) where P(X <= 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) This is what i am getting P(X>3) = -0.00015...............Please let me know if it is correct or no????
Originally Posted by zorro This is what i am getting P(X>3) = -0.00015...............Please let me know if it is correct or no???? Not correct. For starters, a probability cannot be negative. Please show how you got this answer.
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