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

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(52787)   (Show Source): You can put this solution on YOUR website!
.

(i) P = = = 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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    - 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.

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Answer by math_tutor2020(3817)   (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)

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