Question

Let Y be the number of successes in n independent repetitions of a random experiment having probability of success \(p= 2/3\). If \(n= 3\), compute P (\(2 \le Y\)).

• A. \(\frac{20}{27}\)
• B. \(\frac{21}{27}\)
• C. \(\frac{22}{27}\)
• D. \(\frac{23}{27}\)

Hint:

Alternatively, use the complement rule:
\[ P(Y \ge 2) = 1 - [P(Y = 0) + P(Y = 1)] \]
\[ P(Y=0) = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]
\[ P(Y=1) = 3 \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{1}{3}\right)^2 = \frac{6}{27} \]
\[ P(Y \ge 2) = 1 - \frac{7}{27} = \frac{20}{27} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The random variable \(Y\) represents the number of successes in \(n\) independent Bernoulli trials, each with a constant probability of success \(p\).
Therefore, \(Y\) follows a Binomial probability distribution:
\[ Y \sim \text{Binomial}(n, p) \]
Key Formula or Approach:
The probability mass function of a Binomial distribution is:
\[ P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
We want to compute \(P(2 \le Y)\), which is:
\[ P(2 \le Y) = P(Y = 2) + P(Y = 3) \]

Step 2: Detailed Explanation:
Let us identify the parameters from the problem:
- Number of trials, \(n = 3\)
- Probability of success, \(p = \frac{2}{3}\)
- Probability of failure, \(q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}\)
Now, calculate each probability term individually:
- For \(Y = 2\):
\[ P(Y = 2) = \binom{3}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^{3-2} \]
\[ \binom{3}{2} = 3 \]
\[ P(Y = 2) = 3 \cdot \frac{4}{9} \cdot \frac{1}{3} = \frac{12}{27} \]
- For \(Y = 3\):
\[ P(Y = 3) = \binom{3}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^{3-3} \]
\[ \binom{3}{3} = 1 \]
\[ P(Y = 3) = 1 \cdot \frac{8}{27} \cdot 1 = \frac{8}{27} \]
Sum the two probabilities to find the final result:
\[ P(2 \le Y) = P(Y = 2) + P(Y = 3) = \frac{12}{27} + \frac{8}{27} = \frac{20}{27} \]

Step 3: Final Answer:
The correct option is (A).