Consider an experiment of rolling a fair die 5 times. How many elements are in the event space (E3) of getting an outcome where first number
could be (2 or 3) middle number could be either (5,6 or 3) and last number must be 4?
1
Expert's answer
2020-05-24T21:00:33-0400
We have 2 options for first number, 3 options for third number,1 option for last number and 6 options for other numbers. So N=2*6*3*6*1=216 elements
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