Answer:
a. 0.11
b. 0.89
c. 0.33
Step-by-step explanation:
The sample space is an equiprobable sample space since all events have equal probabilities. Let's denote each probability by [tex]x[/tex]. The sum of all probabilities of the events in the sample space is 1.
[tex]9x=1[/tex]
[tex]x=\dfrac{1}{9}=0.111\ldots[/tex]
a. [tex]P(T) = 0.11[/tex] since the events have the same probability.
b. [tex]P(R^c)[/tex] is the probability of [tex]R[/tex] not occurring. Hence, it is the probability of any of the other events.
[tex]P(R^c)=1-0.11=0.89[/tex]
c. [tex]P(P\cup S\cup U)= P(P) + P(S) +P(U) [/tex]
[tex]P(P\cup S\cup U)= \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9}=\frac{1}{3}=0.33[/tex]
Answer:
(a) P(T) ≈ 0.11
(b) P(R n C) = 0
(c) P(P U S U U) ≈ 0.33
Step-by-step explanation:
Given a Sample Space of 9 equally likely events: P, R, S, T, U, V, W, X, and Y. All these events have equal chances of occurrence, they are independent of each other. The probability of occurrence of an event here is simply "number of favourable event ÷ total number of likely event", and that is 1/9.
Now,
(a) Probability of T,
P(T) = 1/9 ≈ 0.11
(b) Probability of R and C,
P(R n C) = 0.
This is because C is not in the Sample Space, and so, it's probability is zero.
Unless you wanted to write P(R n S), and that is P(R) × P(S) = (1/9)×(1/9) = 1/81 ≈ 0.01
(c) Probability of P or S or U, P(P U S U U) = P(P) + P(S) + P(U)
= 1/9 + 1/9 + 1/9
= 3/9
= 1/3
≈ 0.33
