SOLUTION: Using the binomial distribution, find the probability that (i) Three successes [10] in 8 trials when P = 0.4. (ii) Two or fewer successes in 9 trials when P

Algebra ->  Probability-and-statistics -> SOLUTION: Using the binomial distribution, find the probability that (i) Three successes [10] in 8 trials when P = 0.4. (ii) Two or fewer successes in 9 trials when P      Log On


   

Question 1182323: Using the binomial distribution, find the probability that (i) Three successes [10]
in 8 trials when P = 0.4. (ii) Two or fewer successes in 9 trials when P = 0.6
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.


(i)  P = C%5B8%5D%5E3%2A0.4%5E3%2A%281-0.4%29%5E5 = %28%288%2A7%2A6%29%2F%281%2A2%2A3%29%29%2A0.4%5E3%2A0.6%5E5 = 0.2787.    ANSWER



(ii)  It is a binomial distribution probability problem.

    - number of trials         n =   9;
    - number of success trials k <=  2;
    - Probability of success on a single trial p = 0.6.



We need calculate  P(n=9; k <= 2; p=0.6).      


To facilitate calculations, I use an appropriate online (free of charge) calculator at this web-site 

https://stattrek.com/online-calculator/binomial.aspx


It provides nice instructions  and  a convenient input and output for all relevant options/cases.


    P(n=9; k <= 2; p=0.6) = 0.025034752,   or   0.0250 (rounded).       ANSWER

Solved.

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To see a variety of similar solved problems,  look into the lessons
    - Simple and simplest probability problems on Binomial distribution
    - Typical binomial distribution probability problems
    - How to calculate Binomial probabilities with Technology (using MS Excel)
    - Solving problems on Binomial distribution with Technology (using MS Excel)
    - Solving problems on Binomial distribution with Technology (using online solver)
in this site.

After reading these lessons,  you will be able to solve such problems on your own,
which is your  PRIMARY  MAJOR  GOAL  visiting this forum  (I believe).



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Part (i)

n = 8 = number of trials
p = 0.4 = probability of success
k = 3 = number of successes that we want

Binomial probability formula
P(k) = (n C k)*(p)^k*(1-p)^(n-k)
P(3) = (8 C 3)*(0.4)^3*(1-0.4)^(8-3)
P(3) = 56*(0.4)^3*(1-0.4)^(8-3)
P(3) = 0.27869184
This reprsents the probability of getting exactly 3 successes. The decimal value is exact.

Note: the n C k refers to the combination formula but you replace r with k and you get the same idea.

Answer: 0.27869184 (round this however you need to)

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Part (ii)

Use the formula from part (i) to find the following
P(0) = 0.000262144
P(1) = 0.003538944
P(2) = 0.021233664
where n = 9 and p = 0.6 this time (k will range from 0 to 2 as shown above)

Add up those results to get
0.000262144 + 0.003538944 + 0.021233664 = 0.025034752

This represents the probability of getting 2 or fewer (ie at most 2) successes. This decimal value is exact.

Answer: 0.025034752 (round this however you need to)