Question

Suppose that (s, e, p) is a probability space and that e1 and e2 are events satisfying e1 ∪ e2 = s, p(e1) = 1/5, and p(e1 ∩ e2) = 1/30. What is p(e2)?

Answer 1

The  probability space and that e1 and e2 are events satisfying e1 ∪ e2 = s, p(e1) = 1/5, and p(e1 ∩ e2) = 1/30.The  p(e2) is 0.8333.

In chance principle, a probability space or a probability triple. is a mathematical construct that offers a formal model of a random manner or "test". for example, you can still outline an opportunity area that fashions the throwing of a die.

A chance area fashions random occasions and is made of three parts: pattern space: the set of all possible results. as an example, in case you toss a coin twice, the pattern space is {HH, HT, TH, TT}. The sample space is every so often denoted by way of the Greek letter omega (Ω).07-Jan-2017

A finite probability space is a fixed S and a feature p: S → R ≥zero such that p(s) > 0 (∀s ∈ S) and ∑ p(s) = 1. We re. page 1. A finite probability area is a set S and a characteristic p: S → R≥0 such that p(s) > 0. (∀s ∈ S) and ∑

Given E1 and E2 events S is the probability space

And given E1∪ E2 =S

P(E1)=1/5 and P(E1 ∩ E2 )=1/30

From given P(E1∪ E2)=P(S)

From probability axioms P(S)=1

Therefore P(E1∪E2) =1

      P(E1)+ P(E2)- P(E1 ∩E2)= 1

      (1/5) + P(E2) - (1/30) = 1

   P(E2 )= 1-(1/5) +(1/30)= 0.8333

Therefore P(E2)= 0.8333

Learn  more about probability space here brainly.com/question/14596023

#SPJ4