Answer:
n = 3
Step-by-step explanation:
Probability that the packet is received by laptop = 0.8
Probability that the packet is not received by the laptop = 1 - 0.8 = 0.2
The router transmits the message n times. We have to find how many times does the router need to transmit message so that probability that the packet gets through is atleast once is equal to or greater than 0.99
Remember that the event "atleast once" is complement of the event "none"
So,
Event that packets gets through atleast once is complement of event that packets does not get delivered.
Probability that packets doe not get delivered in one attempt = 0.2
Probability that packet does not get delivered in n attempts = [tex](0.2)^{n}[/tex]
So,
The probability that the packet gets delivered atleast once = 1 - Probability that packet does not get delivered in n attempts
The probability that the packet gets delivered atleast once = [tex]1-(0.2)^{n}[/tex]
We need this probability to be greater than or equal to 0.99
So, we can set up the inequality as:
[tex]1-(0.2)^{n}\geq 0.99\\\\ 1-0.99\geq (0.2)^{n}\\\\ 0.01\geq (0.2)^{n}[/tex]
Taking log of both sides, we get:
[tex]\log(0.01)\geq \log((0.2)^{n})\\\\ \log(0.01)\geq n \log(0.2)\\\\ -2\geq -0.699n\\\\\frac{-2}{-0.699} \leq n\\\\ 2.86\leq n[/tex]
The sign of inequality changed because we divided both sides by a negative number.
From the above result we can conclude that n must be equal to or greater than 2.86. Since, n represents the number of times, it cannot be decimal, so the smallest possible value of n would be 3.
Thus, the message must be transmitted atleast 3 times to ensure the probability greater than or equal to 0.99
