Understanding the Probability of Packet Loss in a Congested Network

[francisg3]

a congested computer network has a 0.002 probability of losing a data packet and packet losses are independent events. a lost packet must be resent.

a) what is the probability that an e-mail with 100 packets will need any resent?



i know the answer is P= 1-(1-0.002)^100

however i need some help understanding why it is so. i kno the (1-0.002) si the probability that a packet is not lost. i also know that since there are 100 packets in the e-mail, that there are x^100 chances of it not being lost.

i do not understand why (1-0.002)^100 is subtracted from 1, wouldn't this give the probability that ALL are lost?

also, why can't you just do (probability packets lost)^100?, (0.002)^100

 

[uart]

a congested computer network has a 0.002 probability of losing a data packet and packet losses are independent events. a lost packet must be resent.

a) what is the probability that an e-mail with 100 packets will need any resent?
i know the answer is P= 1-(1-0.002)^100

however i need some help understanding why it is so. i kno the (1-0.002) si the probability that a packet is not lost. i also know that since there are 100 packets in the e-mail, that there are x^100 chances of it not being lost.

i do not understand why (1-0.002)^100 is subtracted from 1, wouldn't this give the probability that ALL are lost?

(1-0.002) is the probability that a packet is NOT lost.

(1-0.002)^100 is the probability that "the first packet is not lost" AND "the second packet is not lost" AND ... AND "the last packet is not lost".

So (1-0.002)^100 is the probability that NO packets are lost. The complement of that is therefore the probability that one or more packets IS lost.

Notice that it's only when events are independent that we are allowed to multiply probabilities like that.

also, why can't you just do (probability packets lost)^100?, (0.002)^100

Ok you haven't thought that one out very well. But perhaps you really meant to ask "why can't we just use (0.002) * 100". That is, why do we have to deal with the complementary probabilities (and then complement the final result) instead of just dealing with it in a simpler manner. Typically the (erroneous) argument goes like this.

Probability of first packet loss = 0.002
Probability of "first packet loss" OR "second packet loss" = 0.002 + 0.002
Probability of "first packet loss" OR "of second packet loss" OR "of third packet loss" = 0.002 + 0.002 + 0.002

...

Probability of 1st OR of second OR of third OR ... OR of last packet loss = 100 * 0.002

You can easily check that the above gives the wrong answer but it's important to understand why. I want you to think about when (under what conditions) that "ANDing" of events corresponds to simple multiplication of their probabilities. Similarly I want you to think about when "ORing" of events corresponds to simple addition of their probabilities. Truly those things are the key to properly understanding this and similar problems.

 

[francisg3]

Thanks for the quick reply!