Four women’s college basketball teams are participating in a single-elimination

Question

Four women’s college basketball teams are participating in a single-elimination holiday basketball tournament. If one team is favored in its semifinal match by odds of 1.95 to 1.05 and another squad is favored in its contest by odds of 2.75 to 1.25, what is the probability that:

a. Both favored teams win their games? (Round your answer to 4 decimal places.)

b. Neither favored team wins its game? (Round your answer to 4 decimal places.)

c. At least one of the favored teams wins its game? (Round your answer to 4 decimal places.)

Explanation / Answer

If the odds is a:b, then the probability of winning is

P = a/(a+b).

The probability that first favored team wins is

P1 = (1.95)/(1.95+1.05) = 0.65

The probability that second favored team wins is

P2 = (2.75)/(2.75+1.25) = 0.6875

a)

P(both favored win) = 0.65*0.6875 = 0.446875 [answer]

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B)

P(neither favored win) = (1-0.65)*(1-0.6875) = 0.109375 [ANSWER]

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c)

P(at least one favored wins) = 1 - P(neither favored wins) = 0.890625 [ANSWER]

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