Answer:
P(X < 3) = 0.7443
Step-by-step explanation:
We are given that the random variable X has a binomial distribution with the given probability of obtaining a success. Also, given n = 6, p = 0.3.
The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 6
r = number of success = less than 3
p = probability of success which in our question is 0.3.
LET X = a random variable
So, it means X ~ 
Now, Probability that X is less than 3 = P(X < 3)
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= 
=
= 0.11765 + 0.30253 + 0.32414 = 0.7443
Therefore, P(X < 3) = 0.7443.
Think rise over run when on a graph
The task is to find the original coordinates with the transformed ones given, so you have to apply the inverse of the stated transformations.
Q"( 6,-1),R"(0,-1) and S"(0,-7)
-> rotate 90 anti-clockwise
Q'(1,6), R'(1,0),S'(7,0)
-> translate left by 7 units
Q(-6,6), R(-6,0), S(0,0)