Use the binomial expression (p+q)^n to calculate a binomial distribution with n = 5 and p= 0.3

0 - 0.036015

1 - 0.36015

2 - 0.3087

3 - 0.1323

4 - 0.02835

5 - 0.00243

Step-by-step explanation:

Honestly don't really know whats going on here but while taking a probability and statistics assignment this question came across and here's what I submitted, answers and work to receive a 100% A binomial distribution is usually a table

(p + q)^5 = p^5 + (5p^4)(q) + (10p^3)(q^2) + (10 p^2)(q^3) + 5p(q^4) + q^5

If p = 0.3, then q = 1 - p = 1 - 0.3 = 0.7

0 - 0.036015

1 - 0.36015

2 - 0.3087

3 - 0.1323

4 - 0.02835

5 - 0.00243

100% I hope this helps

(p + q)^5 = p^5 + (5p^4)(q) + (10p^3)(q^2) + (10 p^2)(q^3) + 5p(q^4) + q^5 

If p = 0.3, then q = 1 - p = 1 - 0.3 = 0.7. 

1. Let X denote the random variable that represents the number of successes. 
2. Let us have values of X:

ranging from no success X=0 to 5 successes X = 5. 

3. Calculate each probability by mutlplying the probability of success ( in this case p = .3) for each success times the probability of failure (1-.3 = .7) for each failure.

4. Then, let us include all the rearrangements for our particular value of X. 
For X= 0, there is no success, so the calculation is .7^5. 
For X=1, we can have 1 success and 4 failures. .3^1 * .7^4, but we can rearrange the successes 5 ways too,

 So, the P(X=1) is 5 * .3^1 * .7^4. 

5. Work out for the rearrangements:

Use the combination rule nCr for each value. 
Then the six calculations are 
.7^5 = .16807 
5C1 .3^1 .7^4 
5C2 .3^2 .7^3 
5C3 .3^3 .7^2 
5C4 .3^4 .7^1 
5C5 .3^5 

6. Using a Ti-84 to do the actual calculations binompdf (5,.3,X) with X ranging from 0 to 5 
It makes up:
.16807 .36015 .3087 .1323  .02835  .00243