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 1.45 to 1.55 and another squad is favored in its contest by odds of 2.25 to 1.75, what is the probability that:

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 1.45 to 1.55 and another squad is favored in its contest by odds of 2.25 to 1.75, what is the probability that:

Explanation / Answer

Odds of 1.45:1.55 implies probability of 1.45/(1.45+1.55) = 29/60 = 0.4833

Odds of 2.65:1.35 implies probability of 2.25/(2.25+1.75) = 9/16 = 0.5625

a)
Probability the two favorites win = 29/60 * 9/16 = 0.2718
b)
Probability that neither favorite wins is = ( 1 - 29/60) * ( 1 - 9/16) = 0.2260
c)
Probability at least 1 favorite wins = 1 - probability neither one wins = 1 - 0.2260 = 0.774

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