Question

For an initial screening of an entrance exam, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is \( \frac{4}{5} \), then the probability that he is unable to solve less than two problems is

• A. $\frac{201}{5}\left(\frac{1}{5}\right)^{49}$
• B. $\frac{316}{25}\left(\frac{4}{5}\right)^{48}$
• C. $\frac{54}{5}\left(\frac{4}{5}\right)^{49}$
• D. $\frac{164}{25}\left(\frac{1}{5}\right)^{48}$

Hint:

Probability Tip:Pay close attention to phrasing: "unable to solve" flips the definitions of $p$ and $q$ from how they are initially presented in the text. "Less than two" strictly means 0 and 1. It does not include 2.

Answer 1

The Correct Option is C

Concept:
Binomial Probability Distribution.

Step 1: Define the binomial distribution parameters.
Let $n = 50$ be the total number of problems. Let $X$ represent the random variable for the number of problems the candidate is unable to solve.

Step 2: Determine the probability of success and failure.
The probability of solving a problem is $q = \frac{4}{5}$. The probability of being unable to solve a problem (which we define as our "success" for variable $X$) is $p = 1 - \frac{4}{5} = \frac{1}{5}$.

Step 3: Set up the mathematical condition based on the question.
We need the probability that he is unable to solve "less than two" problems. This means $X<2$, which translates to finding the sum of probabilities for $X = 0$ and $X = 1$. Thus, we calculate $P(X<2) = P(X=0) + P(X=1)$.

Step 4: Apply the Binomial Probability formula.
Using the formula $P(X=r) = {}^{n}C_{r} p^r q^{n-r}$, we substitute our values:
$P(X<2) = {}^{50}C_{0} \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^{50} + {}^{50}C_{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^{49}$.

Step 5: Simplify the algebraic expression to find the final result.
First, evaluate the combinations: ${}^{50}C_{0} = 1$ and ${}^{50}C_{1} = 50$. Substituting these back gives $1 \cdot 1 \cdot \left(\frac{4}{5}\right)^{50} + 50 \cdot \left(\frac{1}{5}\right) \cdot \left(\frac{4}{5}\right)^{49}$.
We can simplify $50 \cdot \frac{1}{5}$ to $10$.
This leaves us with $\left(\frac{4}{5}\right)^{50} + 10\left(\frac{4}{5}\right)^{49}$.
Next, factor out the common term $\left(\frac{4}{5}\right)^{49}$ from both components to get $\left(\frac{4}{5} + 10\right)\left(\frac{4}{5}\right)^{49}$.
Finally, solve the fraction inside the parentheses: $\frac{4}{5} + \frac{50}{5} = \frac{54}{5}$.
This results in the final formatted answer: $\frac{54}{5}\left(\frac{4}{5}\right)^{49}$. $$ \therefore \text{The required probability is } \frac{54}{5}\left(\frac{4}{5}\right)^{49} $$