A binomial probability experiment is conducted with the given parameters Compute the probability of successes in the independent trals of the experiment

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

Ask by Parsons Parry. in the United States

Mar 26,2025

Question

A binomial probability experiment is conducted with the given parameters Compute the probability of successes in the independent trals of the experiment

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

Ask by Parsons Parry. in the United States

Mar 26,2025

UpStudy AI · Accepted Answer

We start with the formula for the probability of exactly $$ x $$ successes in $$ n $$ trials:

$$
P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}
$$

For $$ n=9 $$ and $$ p=0.5 $$, the probability of an exact outcome of $$ x $$ successes is:

$$
P(X = x) = \binom{9}{x}(0.5)^x(0.5)^{9-x} = \binom{9}{x}(0.5)^9
$$

We need the probability of having at most 3 successes ($$ x \leq 3 $$), so we sum from $$ x=0 $$ to $$ x=3 $$:

$$
P(X \leq 3) = \sum_{x=0}^{3} \binom{9}{x}(0.5)^9
$$

Factor out the constant $$(0.5)^9$$:

$$
P(X \leq 3) = (0.5)^9 \sum_{x=0}^{3} \binom{9}{x}
$$

Next, we compute the binomial coefficients:

$$
\binom{9}{0} = 1,\quad \binom{9}{1} = 9,\quad \binom{9}{2} = 36,\quad \binom{9}{3} = 84
$$

Adding these up gives:

$$
\sum_{x=0}^{3} \binom{9}{x} = 1 + 9 + 36 + 84 = 130
$$

Since $$(0.5)^9= \frac{1}{512}$$, we have:

$$
P(X \leq 3) = 130 \cdot \frac{1}{512} = \frac{130}{512} = \frac{65}{256}
$$

Evaluating numerically:

$$
\frac{65}{256} \approx 0.2539
$$

Thus, the probability of at most 3 successes is:

$$
\boxed{0.2539}
$$
The probability of at most 3 successes is 0.2539.