Consider the trial on which a 3 is first observed in successive rolls of a six-sided die. let a be the event that 3 is observed on the first trial. let b be the event that at 3.assumingthat each side has probability 1/6,

1. p(a)
2. p(b)
3.p(a u b)

1) 0.1667

2) 0.8333

3) 1

Step-by-step explanation:

Let A be the event that 3 is observed on the first trial.

Let B be the event that at least two trials are required to observe a 3

a) A six-sided die has 6 possible outcomes of 1,2,3,4,5,6

3 is one of the six possible outcomes

P(A) = no of favourable outcomes /no of possible outcomes

P(A) = 1/6 = 0.1667

b) The event of requiring at least two trials to observe a 3 is the event of not observing a 3 on the first trial.

P(B) = P(A')

P(A')= 1 - P(A)

P(B) = 1 - 1/6

= 5/6 = 0.8333

c) P(AnB) = P(A) + P(B)

= 1/6 + 5/6

= 6/6 = 1

30005

yes ok