Apacket processed by an internet router carries either audio information with probability 7/10, video, with probability 2/10, or text with probability 1/10. the probability that the router processes 70 audio packets, 20 video packets, and 10 text packets in a sequence of 100 packets is:

An interesting problem using hypergeometric distribution.

Here we have three variables, 
a=number of audio packets
v=number of video packets
t=number of text packets
we note that a+v+t=100 (as given)

and correspondingly, 
A=number of audio packets in the pool
V=number of video packets in the pool
T=number of text packets in the pool
Here, the pool is from a population such that the probabilities of selecting each type are respectively 0.7, 0.2 and 0.1  (as given).

Defining C(n,r)=n!/(r!(n-r)!) as the number of combinations of r objects chosen out of n.

In a pool of 1000 packets, then A=700, V=200 and T=100.
The multivariate hypergeometric distribution gives
P(70,20,10,1000)
=C(700,70)*C(200,20)*C(100,10)/C(1000,100)
=0.0147526

In a pool of 100000 packets, then A=70000,V=20000,T=10000.
The multivariate hypergeometric distribution gives
P(70,20,10,1000)
=C(70000,70)*C(20000,20)*C(10000,10)/C(100000,100)
=0.0132925

Since the population has a size of infinite, we will try to approach the value by choosing an even larger population, say, 1000000,
P(70,20,10,1000000)=0.0132806

Next try
P(70,20,10,1000000000)=0.013279255
P(70,20,10,1000000000000)=0.013279254
...
So we can conclude that
P(70,20,10,population)=0.013279 with almost certainty.

Point D

Step-by-step explanation: Now you want to look at the point closest to -0.4. Think about it. The only other possible point could be C. But thats not it because C is kinda in the middle of -0.4 and -0.5