Understanding the Problem

We have 5 independent Bernoulli trials, each with probability of success $p$. The probability of at least one failure is given to be greater than or equal to $\frac{31}{32}$. We need to find the interval in which $p$ lies.

Key Concept: Complementary Probability

The probability of "at least one failure" is the complement of the event "no failures" (i.e., all successes).

Let $P\left(A\right)$ = Probability of at least one failure.

Let $P\left(B\right)$ = Probability of all successes.

Therefore, $P\left(A\right)=1-P\left(B\right)$.

Since the trials are independent, the probability of all successes is $p^5$.

Step-by-Step Solution

Step 1: Write the inequality based on the given condition.

Probability of at least one failure $\ge \frac{31}{32}$.

This means: $1-p^5\ge \frac{31}{32}$.

Step 2: Solve the inequality for $p^5$.

Subtract 1 from both sides and multiply by -1 (remembering to reverse the inequality sign):

$1-p^5\ge \frac{31}{32}$

$\Rightarrow -p^5\ge \frac{31}{32}-1$

$\Rightarrow -p^5\ge -\frac{1}{32}$

$\Rightarrow p^5\le \frac{1}{32}$ (Multiplying both sides by -1 reverses the inequality)

Step 3: Take the fifth root of both sides to solve for $p$.

$p\le {\left(\frac{1}{32}\right)}^{1/5}$

$p\le {\left(\frac{1}{2^5}\right)}^{1/5}$

$p\le \frac{1}{2}$

Step 4: Consider the domain of $p$.

Since $p$ is a probability, it must lie between 0 and 1, inclusive: $0\le p\le 1$.

Our inequality from Step 3 gives us $p\le \frac{1}{2}$.

Combining this with the domain, we get the final interval: $0\le p\le \frac{1}{2}$.

Final Answer: $p\in \left[0,\frac{1}{2}\right]$

This corresponds to the option: $\left[0,\frac{1}{2}\right]$

Related Topics & Formulae

Bernoulli Trials: A random experiment with exactly two possible outcomes, "success" and "failure".

Binomial Distribution: The probability of getting exactly $k$ successes in $n$ independent Bernoulli trials is given by: $P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{(1-p)}^{n-k}$

Complementary Probability: For any event $A$, the probability that $A$ occurs is $1-P\left(\text{not }A\right)$. This is often used to simplify problems involving "at least one" occurrence of an event.