Question

One die is thrown two times. If the sum of the appeared numbers on their faces is 6, find the conditional probability of appearing number 4 at least one time in that.

Hint:

Conditional Probability = $\dfrac{n(\text{Favourable outcomes satisfying condition})}{n(\text{Total outcomes under given condition})}$.

Answer 1

Step 1: Total possible outcomes. When two dice are thrown, total outcomes = $36$. Given condition: sum of numbers = 6.

Step 2: Outcomes where sum = 6. \[ (1,5), (2,4), (3,3), (4,2), (5,1) \] So, total favourable cases for sum = 6 → $5$.

Step 3: Outcomes where at least one 4 appears (within sum = 6). \[ (2,4), (4,2) \] So, number of favourable outcomes = $2$.

Step 4: Conditional probability. \[ P(\text{at least one 4} \mid \text{sum = 6}) = \frac{\text{Favourable outcomes}}{\text{Total outcomes with sum = 6}} = \frac{2}{5} \]

Final Answer: \[ \boxed{\dfrac{2}{5}} \]