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)
To determine the elements of the event A (the team wins at least twice and does not lose), we can analyze the different possible outcomes for the three games.
Since the team plays three games, there are 2^3 = 8 possible outcomes, which we can represent as combinations of W (Win), L (Loss), and T (Tie).
The elements of event A are the outcomes that satisfy the given conditions: the team wins at least twice and does not lose (more than two wins and zero losses).
Let's go through each possible outcome and determine if it belongs to event A:
1. WWW: This outcome satisfies both conditions (more than two wins and zero losses).
2. WWL: This outcome satisfies the first condition but not the second (one loss).
3. WLW: This outcome satisfies only the first condition (two wins and one loss).
4. WLL: This outcome satisfies neither condition (one win and two losses).
5. LWW: This outcome satisfies only the first condition (two wins and one loss).
6. LWL: This outcome satisfies neither condition (one win and two losses).
7. LLW: This outcome satisfies neither condition (one win and two losses).
8. LLL: This outcome satisfies neither condition (zero wins and three losses).
So, the elements of event A are {WWW, WLW, LWW}.
To find the probability of event A (P(A)), we need to calculate the probability of each element in A and sum them up.
P(A) = P(WWW) + P(WLW) + P(LWW)
Given that P(W) = 0.6, P(L) = 0.3, and P(T) = 0.1, we can calculate the probabilities for each outcome:
P(WWW) = P(W) * P(W) * P(W) = 0.6 * 0.6 * 0.6 = 0.216
P(WLW) = P(W) * P(L) * P(W) = 0.6 * 0.3 * 0.6 = 0.108
P(LWW) = P(L) * P(W) * P(W) = 0.3 * 0.6 * 0.6 = 0.108
P(A) = 0.216 + 0.108 + 0.108 = 0.432
Therefore, the probability of event A (the team wins at least twice and does not lose) is 0.432.
Moving on to event B (the team wins, loses, and ties), we need to find the different outcomes that satisfy this condition.
To win, lose, and tie in three games, the only possible outcome is WLT.
Therefore, the elements of event B are {WLT}.
And the probability of event B (P(B)) is simply the probability of the outcome WLT:
P(B) = P(W) * P(L) * P(T) = 0.6 * 0.3 * 0.1 = 0.018
So, the probability of event B (the team wins, loses, and ties) is 0.018.