Answer:
4/9
Step-by-step explanation:
Given that :
Die is designed such that even numbers are twice as likely to occur as odd number.
Hence,
Sample space :
1, 2, 2, 3, 4, 4, 5, 6, 6
Total possible outcomes = 9
Probability that a number less than 4 occurs on a single toss of the die.
Probability = required outcome / Total possible outcomes
Required outcome = number less than 4 [1, 2, 2, 3]
Hence,
P(obtaining a number less than 4):
(Number of possible outcomes less than 4 / total number of events in sample space)
= 4 / 9
