Question

A die is tossed twice. If a “success” is getting a number greater than 4, find the probability distribution of the number of successes.

Hint:

Repeated independent trials with two outcomes follow a binomial distribution. Identify \( n \) and \( p \) first.

Answer 1

Concept: A success is getting a number greater than 4 (i.e., 5 or 6). So probability of success: \[ P(S) = \frac{2}{6} = \frac{1}{3} \] Probability of failure: \[ P(F) = \frac{4}{6} = \frac{2}{3} \] Since the die is tossed twice independently, this is a binomial distribution with: \[ n = 2, \quad p = \frac{1}{3} \]
Step 1: Let \( X \) = number of successes in two tosses. Possible values: \( X = 0, 1, 2 \)
Step 2: Use binomial probability formula \[ P(X = k) = {2 \choose k} p^k (1-p)^{2-k} \] For X = 0: \[ P(0) = (1-p)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \] For X = 1: \[ P(1) = 2 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{9} \] For X = 2: \[ P(2) = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]
Step 3: Probability distribution \[ \begin{array}{c|c} X & P(X)
\hline 0 & \frac{4}{9}
1 & \frac{4}{9}
2 & \frac{1}{9} \end{array} \] Final Answer: The probability distribution is: \[ P(0)=\frac{4}{9}, \quad P(1)=\frac{4}{9}, \quad P(2)=\frac{1}{9} \] Explanation: Each toss is independent with success probability \( \frac{1}{3} \). Hence, the number of successes follows a binomial distribution.