Understanding the Problem

This is a binomial probability problem. We have 5 independent questions, each with 3 choices, and exactly one is correct. The probability of guessing a single question correctly is $p=\frac{1}{3}$, and the probability of guessing incorrectly is $q=1-p=\frac{2}{3}$. We need to find the probability of getting 4 or more correct answers, which means exactly 4 correct or exactly 5 correct.

Step-by-Step Solution

Step 1: Define the Binomial Probability Formula

The probability of getting exactly $k$ successes in $n$ trials is given by:

$P(X=k)=C(n,k)\cdot p^k\cdot q^{n-k}$

Where $C(n,k)=\frac{n!}{k!(n-k)!}$ is the number of combinations.

Step 2: Calculate Probability for Exactly 4 Correct Answers (k=4)

Here, $n=5$, $k=4$, $p=\frac{1}{3}$, $q=\frac{2}{3}$.

First, find the number of ways to choose 4 correct answers out of 5:

$C(5,4)=\frac{5!}{4!(5-4)!}=\frac{5\cdot 4!}{4!\cdot 1!}=5$

Now, calculate the probability:

$P(X=4)=5\cdot {\left(\frac{1}{3}\right)}^4\cdot {\left(\frac{2}{3}\right)}^1=5\cdot \frac{1}{3^4}\cdot \frac{2}{3}=5\cdot \frac{1}{81}\cdot \frac{2}{3}=\frac{10}{3^5}$

(Note: $3^4\cdot 3=3^5=243$)

Step 3: Calculate Probability for Exactly 5 Correct Answers (k=5)

Here, $n=5$, $k=5$, $p=\frac{1}{3}$, $q=\frac{2}{3}$.

First, find the number of ways to choose 5 correct answers out of 5:

$C(5,5)=1$

Now, calculate the probability:

$P(X=5)=1\cdot {\left(\frac{1}{3}\right)}^5\cdot {\left(\frac{2}{3}\right)}^0=1\cdot \frac{1}{3^5}\cdot 1=\frac{1}{3^5}$

Step 4: Find the Total Probability (4 or more correct)

Since the events are mutually exclusive, we add the two probabilities:

$P(X\ge 4)=P(X=4)+P(X=5)=\frac{10}{3^5}+\frac{1}{3^5}=\frac{11}{3^5}$

Final Answer: The probability that a student will get 4 or more correct answers just by guessing is $\frac{11}{3^5}$.

Related Topics & Formulae

Binomial Distribution: A probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters. The parameters are the number of trials ($n$) and the probability of success ($p$) in a single trial.

Key Formula:

$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$

Where $q=1-p$ and $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$

Cumulative Probability: To find the probability of a range of outcomes (e.g., $X\ge k$), you sum the probabilities of all individual outcomes in that range.