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15 August, 13:25

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?

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  1. 15 August, 13:42
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    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
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