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

Given: 
A die is tossed twice.
A “success” is defined as getting a number greater than 4.

Step 1: Find Probability of Success in One Trial
Numbers greater than 4 on a die are: 5 and 6.

So, favourable outcomes = 2
Total possible outcomes = 6

Probability of success (p) = 2/6 = 1/3

Probability of failure (q) = 1 − p = 1 − 1/3 = 2/3

Step 2: Define Random Variable
Let X = number of successes in two tosses.

Since the die is tossed twice, possible values of X are:

X = 0, 1, 2

This follows Binomial Distribution with:
n = 2, p = 1/3

Step 3: Calculate Probabilities

P(X = 0) = C(2,0) (1/3)⁰ (2/3)²
= 1 × 1 × 4/9 = 4/9

P(X = 1) = C(2,1) (1/3)¹ (2/3)¹
= 2 × (1/3) × (2/3) = 4/9

P(X = 2) = C(2,2) (1/3)² (2/3)⁰
= 1 × 1/9 × 1 = 1/9

Step 4: Probability Distribution Table

X (Number of Successes)	P(X)
0	4/9
1	4/9
2	1/9

Conclusion:
The probability distribution of the number of successes is:

P(0) = 4/9,   P(1) = 4/9,   P(2) = 1/9