Math Problem Statement

Suppose a random variable, x, arises from a binomial experiment. If n = 10, and p = 0.1, find the following probabilities using the binomial formula. Round to four decimal places, if necessary.

P(x = 9) = 

P(x = 5) = 

P(x = 2) = 

P(x ≤ 1) = 

P(x ≥ 8) = 

P(x ≤ 7) = 

Solution

To solve these binomial probabilities, we will use the binomial probability formula:

P(x=k)=(nk)pk(1−p)n−kP(x = k) = \binom{n}{k} p^k (1 - p)^{n - k}

where:

  • n=10n = 10 (the number of trials),
  • p=0.1p = 0.1 (the probability of success),
  • 1−p=0.91 - p = 0.9 (the probability of failure),
  • kk is the number of successes we want to find the probability for,
  • (nk)\binom{n}{k} is the binomial coefficient, calculated as n!k!(n−k)!\frac{n!}{k!(n-k)!}.

Now, we calculate each of the requested probabilities:

1. P(x=9)P(x = 9):

= \frac{10!}{9!(10-9)!} (0.1)^9 (0.9) = 10 \cdot (0.1)^9 \cdot 0.9$$ $$P(x = 9) = 10 \cdot 0.000000001 \cdot 0.9 = 0.000000009$$ Rounded to four decimal places: $$P(x = 9) = 0.0000$$. ### 2. $$P(x = 5)$$: $$P(x = 5) = \binom{10}{5} (0.1)^5 (0.9)^5 = \frac{10!}{5!(10-5)!} (0.1)^5 (0.9)^5 = 252 \cdot (0.1)^5 \cdot (0.9)^5$$ $$P(x = 5) = 252 \cdot 0.00001 \cdot 0.59049 = 0.001488$$ So, $$P(x = 5) = 0.0015$$. ### 3. $$P(x = 2)$$: $$P(x = 2) = \binom{10}{2} (0.1)^2 (0.9)^8 = \frac{10!}{2!(10-2)!} (0.1)^2 (0.9)^8 = 45 \cdot (0.1)^2 \cdot (0.9)^8$$ $$P(x = 2) = 45 \cdot 0.01 \cdot 0.43046721 = 0.19371$$ So, $$P(x = 2) = 0.1937$$. ### 4. $$P(x \leq 1)$$: This represents the cumulative probability of $$P(x = 0)$$ and $$P(x = 1)$$. #### $$P(x = 0)$$: $$P(x = 0) = \binom{10}{0} (0.1)^0 (0.9)^{10} = 1 \cdot 1 \cdot (0.9)^{10} = 0.3487$$ #### $$P(x = 1)$$: $$P(x = 1) = \binom{10}{1} (0.1)^1 (0.9)^9 = 10 \cdot 0.1 \cdot (0.9)^9 = 0.3874$$ Now, sum the two probabilities: $$P(x \leq 1) = P(x = 0) + P(x = 1) = 0.3487 + 0.3874 = 0.7361$$ ### 5. $$P(x \geq 8)$$: This represents the cumulative probability of $$P(x = 8)$$, $$P(x = 9)$$, and $$P(x = 10)$$. #### $$P(x = 8)$$: $$P(x = 8) = \binom{10}{8} (0.1)^8 (0.9)^2 = 45 \cdot 0.00000001 \cdot 0.81 = 0.00000036$$ #### $$P(x = 10)$$: $$P(x = 10) = \binom{10}{10} (0.1)^{10} (0.9)^0 = 1 \cdot 0.0000000001 = 0.0000000001$$ Now, sum the probabilities for $$x = 8$$, $$x = 9$$, and $$x = 10$$: $$P(x \geq 8) = 0.00000036 + 0.000000009 + 0.0000000001 = 0.0000003691$$ So, rounded to four decimal places: $$P(x \geq 8) = 0.0000$$. ### 6. $$P(x \leq 7)$$: This is the complement of $$P(x \geq 8)$$: $$P(x \leq 7) = 1 - P(x \geq 8) = 1 - 0.0000 = 1.0000$$ ### Final Answers: 1. $$P(x = 9) = 0.0000$$ 2. $$P(x = 5) = 0.0015$$ 3. $$P(x = 2) = 0.1937$$ 4. $$P(x \leq 1) = 0.7361$$ 5. $$P(x \geq 8) = 0.0000$$ 6. $$P(x \leq 7) = 1.0000$$ Would you like more details or have any other questions? Here are 5 related questions you might explore: 1. What is the expected value and variance for this binomial distribution? 2. How does increasing $$p$$ affect the shape of the binomial distribution? 3. Can you calculate $$P(2 \leq x \leq 5)$$ for this distribution? 4. What happens to the probabilities if $$n$$ is increased to 20? 5. How would you approximate these probabilities using a normal distribution? **Tip**: The binomial distribution can be approximated by a normal distribution when $$n$$ is large and $$p$$ is not too close to 0 or 1.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution

Formulas

P(x = k) = C(n, k) * p^k * (1 - p)^(n - k)
C(n, k) = n! / (k!(n-k)!)

Theorems

Binomial Theorem

Suitable Grade Level

Grades 9-12

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