given that z ~B(20,0.3), find the probability that?

Question

z is less than or equal to 2, given that z is less than or equal to 4.

Answer 1

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.

the mean of the binomial distribution is n * p
the variance of the binomial distribution is n * p * (1 - p)

for any two events A and B, with P(B) ≠ 0, then the conditional probability of A given B is:

P(A | B) = P( A ∩ B) / P(B)

for your problem: you have P(Z ≤ 2 | Z ≤4)

= P(Z ≤ 2 ∩ Z ≤ 4) / P(Z ≤ 4)
= P(Z ≤ 2 ) / P(Z ≤ 4) because if Z is ≤ 2 it is also ≤ 4

P(Z ≤ 2) = P(Z = 0) + P(Z = 1) + P(Z = 2)
= 0.0007979227 + 0.0068393371 + 0.0278458725
= 0.03548313

P(Z ≤ 4) = P(Z = 0) + P(Z = 1) + P(Z = 2) + P(Z = 3) + P(Z = 4)
= 0.0007979227 + 0.0068393371 + 0.0278458725 + 0.0716036722 + 0.1304209744
= 0.2375078

so

P(Z ≤ 2 | Z ≤4)
= P(Z ≤ 2 ) / P(Z ≤ 4)
= 0.03548313 / 0.2375078
= 0.1493977

Answer 2

using the binomial formula (which I will not try to type here!) from your textbook, you can find

P (z <= 4) = P(z=0) + P(z=1) + P(z=2) + P(z=3) + P(z=4)
P (z <= 2) = P(z=0) + P(z=1) + P(z=2)

now
P ( z <= 2 | z <= 4) = P( z <= 2 and z <= 4)/ P( z <=4)

= P( z <= 2 )/ P( z <=4)

(since the two numerators are describing the same event!)
So just take the ratio of the two binomial probabilities computed on the first two lines.

Answer 3

You have an answer. You can use the source to compute your probabilities.
n=20, p=0.3