Three friends (A, B, and C) will participate in a round-robin tournament in whic
ID: 3199965 • Letter: T
Question
Three friends (A, B, and C) will participate in a round-robin tournament in which each one plays both of the others. Suppose that
P(A beats B) = 0.9
P(A beats C) = 0.7
P(B beats C) = 0.5
and that the outcomes of the three matches are independent of one another.
(a) What is the probability that A wins both her matches and that B beats C?
(b) What is the probability that A wins both her matches?
(c) What is the probability that A loses both her matches?
(d) What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)
There are two traffic lights on the route used by a certain individual to go from home to work. Let E denote the event that the individual must stop at the first light, and define the event F in a similar manner for the second light. Suppose that P(E) = 0.4, P(F) = 0.2, and P(E F) = 0.15.
(a) What is the probability that the individual must stop at at least one light; that is, what is the probability of the event P(E F)?
(b) What is the probability that the individual doesn't have to stop at either light?
(c) What is the probability that the individual must stop at exactly one of the two lights?
(d) What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to P(E) and P(E F)? A Venn diagram might help.)
** Please can i get answers and solutions please? I don't seem to understand this, please help. Thank you.
Explanation / Answer
1 Solution:-
given data P(A beats B) = 0.9
P(A beats C) = 0.7
P(B beats C) = 0.5
a) 0.9 * 0.7 * .5 = 0.315
b) 0.9 * 0.7 = 0.63
c) 0.7 * 0.9 = 0.63
d) This can happen two ways.
A beats B, B beats C, C beats A or
A beats C, C beats B, B beats A
(0.9 * 0.7 * 0.9) + (0.1 * 0.3 * 0.1) = 0.567 + 0.003 = 0.57
2 solution:-
given data P(E) = 0.4, P(F) = 0.2, and P(E F) = 0.15.
a) P(E F) = P(E) + P(F) - P(E F)
= 0.4 + 0.2 - 0.15
= 0.45
b) 1- P(E F) = 1- 0.45 = 0.55
c) P( E F') + P( E' F) = (0.4)(1-0.2) + (1-0.4)(0.2) = 0.44
d) P(E) = P( E F) + P ( E F') = (0.4)(0.2) + (0.4)(1-0.2)
= 0.08 + 0.32
= 0.40