A certain soccer team wins (W) with P(W)= 60%, loses(L) with P(L) = 30%, and ties (T) with P(T) = 10%. The team plays three games over the weekend.
a.Determine the elements of the event A that the team wins at least twice and does not lose.
b.Find P(A)
c.Determine the elements of the event B that the team wins, loses, and ties
d.Find (B)
a) Could be
WWW
WWT -- 3 ways for that
b) P(A) = .6^3 + 3(.6)^2 (.1) = ...
c) WLT - 6 ways to arrange that
d) P)B) = 6(.6)(.3)(.1) = ....
you do the button pushing
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To determine the elements of the given events and calculate their probabilities, we can use basic probability rules and combinatorics.
Let's start with part a.
a. The event A represents the scenario where the team wins at least twice and does not lose. There are a few possible outcomes for this event:
1. Win-Win-Win (WWW)
2. Win-Win-Tie (WWT)
3. Win-Tie-Win (WTW)
4. Tie-Win-Win (TWW)
Notice that we do not count the scenario where the team loses any game.
Next, let's calculate the probabilities for each outcome:
1. P(WWW) = P(W) * P(W) * P(W) = 0.60 * 0.60 * 0.60 = 0.216
2. P(WWT) = P(W) * P(W) * P(T) = 0.60 * 0.60 * 0.10 = 0.036
3. P(WTW) = P(W) * P(T) * P(W) = 0.60 * 0.10 * 0.60 = 0.036
4. P(TWW) = P(T) * P(W) * P(W) = 0.10 * 0.60 * 0.60 = 0.036
The elements of event A are WWW, WWT, WTW, and TWW.
b. To find the probability of event A, we need to sum up the probabilities of all the individual outcomes:
P(A) = P(WWW) + P(WWT) + P(WTW) + P(TWW) = 0.216 + 0.036 + 0.036 + 0.036 = 0.324
c. The event B represents the scenario where the team wins, loses, and ties. There is only one outcome for this event: Win-Loss-Tie (WLT).
Let's calculate the probability for this outcome:
P(WLT) = P(W) * P(L) * P(T) = 0.60 * 0.30 * 0.10 = 0.018
The elements of event B are WLT.
d. The probability of event B is simply the probability of the outcome WLT:
P(B) = P(WLT) = 0.018
Therefore, the elements of event B are WLT and the probability of event B is 0.018.