Question

A dice is thrown twice. What is the probability that: i) 6 will not come up either time? ii) 6 will come up at least once?

Hint:

$P(\text{at least once}) = 1 - P(\text{not even once})$. This makes part (ii) very easy once you have part (i).

Answer 1

Step 1: Understanding the Concept:
When a die is thrown twice, the total number of outcomes is $6 \times 6 = 36$.
Step 2: Probability calculation for (i):
Let $E$ be the event where '6' comes up at least once. Outcomes are:
$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (1,6), (2,6), (3,6), (4,6), (5,6)$. Total outcomes for 'at least one 6' = 11.
Outcomes where '6' does not come up either time = $36 - 11 = 25$. \[ P(\text{no 6}) = \frac{25}{36} \]
Step 3: Probability calculation for (ii):
'6 will come up at least once' is the set we just identified above. Number of favorable outcomes = 11. \[ P(\text{at least one 6}) = \frac{11}{36} \]
Step 4: Final Answer:
The probabilities are 25/36 and 11/36 respectively.