A congested computer network has a 0.005 probability of losing a data packet and packet losses are independent events. A lost packet must be resent. Round your answers to four decimal places (e. g. 98.7654).

a. What is the probability that an e-mail message with 100 packets will need any resent?
b. What is the probability that an e-mail message with 3 packets will need exactly one to be resent?
c. If 10 e-mail messages are sent, each with 100 packets, what is the probability that at least one message will need some packets to be resent?

Answer

given,

Probability of loosing data, p = 0.005

 q = 1 - 0.005 = 0.995

a) Probability that none of the need resent in the 100 Pack

  = ¹⁰⁰C₀ p⁰q¹⁰⁰

 = 0.995¹⁰⁰ = 0.6057

 Probability at least one mail is resent = 1 - 0.6057

                                                                = 0.3942

b) Probability that that exactly one out of three needs to be resent is:

   =  ³C₁ p¹q²

   =  3 x 0.005 x 0.995²

   = 0.0149

c) Probability that none of the need resent in the 1000 Pack

  = ¹⁰⁰⁰C₀ p⁰q¹⁰⁰⁰

 = 0.995¹⁰⁰⁰ = 0.006657

 Probability at least one mail is resent = 1 - 0.006657

                                                                = 0.9933

answer:

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explanation: