A binomial probability experiment is conducted with the given parameters. Compute the probabilify of successes in the independenf frials of the experiment.

The probability of successes is . (Round to four decimal places as needed.)
(Ry…

Ask by Mckenzie Guerrero. in the United States

Mar 26,2025

Question

A binomial probability experiment is conducted with the given parameters. Compute the probabilify of successes in the independenf frials of the experiment.

The probability of successes is . (Round to four decimal places as needed.)
(Ry…

Ask by Mckenzie Guerrero. in the United States

Mar 26,2025

UpStudy AI · Accepted Answer

Given a binomial experiment with parameters  
$$
n = 12, \quad p = 0.1,
$$  
we wish to compute the probability of at most 4 successes, i.e.,  
$$
P(X \le 4) = \sum_{x=0}^{4} \binom{12}{x} (0.1)^x (0.9)^{12-x}.
$$

**Step 1. Write the expression for each term**

For a given $$ x $$, the probability is  
$$
P(X = x) = \binom{12}{x}(0.1)^x(0.9)^{12-x}.
$$

Thus,  
$$
P(X \le 4) = \binom{12}{0}(0.1)^0(0.9)^{12} + \binom{12}{1}(0.1)^1(0.9)^{11} + \binom{12}{2}(0.1)^2(0.9)^{10} + \binom{12}{3}(0.1)^3(0.9)^9 + \binom{12}{4}(0.1)^4(0.9)^8.
$$

**Step 2. Compute each term**

1. For $$ x = 0 $$:
   $$
   P(X = 0) = \binom{12}{0}(0.1)^0(0.9)^{12} = 1 \cdot 1 \cdot (0.9)^{12}.
   $$
   Calculating,  
   $$
   (0.9)^{12} \approx 0.2824.
   $$

2. For $$ x = 1 $$:
   $$
   P(X = 1) = \binom{12}{1}(0.1)^1(0.9)^{11} = 12 \cdot 0.1 \cdot (0.9)^{11}.
   $$
   Since  
   $$
   (0.9)^{11} \approx 0.3138,
   $$
   then  
   $$
   P(X = 1) \approx 12 \cdot 0.1 \cdot 0.3138 = 0.3766.
   $$

3. For $$ x = 2 $$:
   $$
   P(X = 2) = \binom{12}{2}(0.1)^2(0.9)^{10} = 66 \cdot 0.01 \cdot (0.9)^{10}.
   $$
   With  
   $$
   (0.9)^{10} \approx 0.3487,
   $$
   we have  
   $$
   P(X = 2) \approx 66 \cdot 0.01 \cdot 0.3487 \approx 0.2300.
   $$

4. For $$ x = 3 $$:
   $$
   P(X = 3) = \binom{12}{3}(0.1)^3(0.9)^9 = 220 \cdot 0.001 \cdot (0.9)^9.
   $$
   Where  
   $$
   (0.9)^9 \approx 0.3874,
   $$
   so  
   $$
   P(X = 3) \approx 220 \cdot 0.001 \cdot 0.3874 \approx 0.0852.
   $$

5. For $$ x = 4 $$:
   $$
   P(X = 4) = \binom{12}{4}(0.1)^4(0.9)^8 = 495 \cdot 0.0001 \cdot (0.9)^8.
   $$
   With  
   $$
   (0.9)^8 \approx 0.4305,
   $$
   it follows that  
   $$
   P(X = 4) \approx 495 \cdot 0.0001 \cdot 0.4305 \approx 0.0213.
   $$

**Step 3. Sum the probabilities**

Now,  
$$
\begin{aligned}
P(X \le 4) &\approx 0.2824 + 0.3766 + 0.2300 + 0.0852 + 0.0213 \
           &\approx 0.9945.
\end{aligned}
$$

Thus, the probability of at most 4 successes is  
$$
\boxed{0.9945}.
$$
The probability of at most 4 successes is approximately 0.9945.

A binomial probability experiment is conducted with the given parameters. Compute the probabilify of successes in the independenf frials of the experiment.

The probability of successes is . (Round to four decimal places as needed.)
(Ry…

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A binomial probability experiment is conducted with the given parameters. Compute the probabilify of successes in the independenf frials of the experiment. The probability of successes is . (Round to four decimal places as needed.) (Ry…

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