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In Exercises \(1-10,\) you are performing five independent Bernoulli trials with \(p=.1\) and \(q=.9 .\) Calculate the probability of the stated outcome. Check your answer using technology. [HINT: See Quick Example \(8 .\) At most three successes

Short Answer

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The probability of at most three successes in five independent Bernoulli trials with \(p=0.1\) and \(q=0.9\) is approximately 0.99954.
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Step 1: Probabilities for each outcome

Suppose the number of successes is denoted by k. We need to calculate the probabilities of each outcome, i.e., \(k=0,1,2,3\). For each k, we use the binomial probability formula: \[P(k)=\binom{n}{k} p^k q^{n-k}\] where n is the number of trials, p is the probability of success, and q is the probability of failure.

Step 2: 0 Successes

For 0 successes, we plug in \(k=0\). \[P(0)=\binom{5}{0} \times 0.1^{0} \times 0.9^{5-0}=1 \times 1\times 0.9^5=0.59049\]

Step 3: 1 Success

For 1 success, we plug in \(k=1\). \[P(1)=\binom{5}{1} \times 0.1^{1} \times 0.9^{5-1}=5 \times 0.1\times 0.9^4=0.32805\]

Step 4: 2 Successes

For 2 successes, we plug in \(k=2\). \[P(2)=\binom{5}{2} \times 0.1^{2} \times 0.9^{5-2}=10 \times 0.1^2\times 0.9^3=0.0729\]

Step 5: 3 Successes

For 3 successes, we plug in \(k=3\). \[P(3)=\binom{5}{3} \times 0.1^{3} \times 0.9^{5-3}=10 \times 0.1^3\times 0.9^2=0.0081\]

Step 6: Summing the Probabilities

Now we sum up the probabilities for each outcome. \[P(\text{At most 3 successes})=P(0)+P(1)+P(2)+P(3)=0.59049+0.32805+0.0729+0.0081=0.99954\] The probability of at most three successes is approximately 0.99954.

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