Q16. (a) Bag \( A \) contains 3 blue balls and 7 green balls. Bag \( B \) contains 5 blue balls and 9 green balls. A ball is taken from bag \( A \) and placed in bag \( B \). One ball is then taken from bag \( B \). Given that the selections are random, find the probability that the ball taken from bag \( B \) is green. (b) The probabilities that Team \( M \) wins and loses a soccer match against Team \( L \) are \( \frac{8}{15} \) and \( \frac{1}{3} \) respectively. If there are two soccer matches between the two teams, calculate (i) the probability that Team \( M \) wins one game and loses one game against Team \( L \). (ii) the probability that Team \( M \) does not lose in both matches played against Team \( L \).
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To find the probability of selecting a green ball from bag \( B \) after transferring a ball from bag \( A \), we can analyze the scenarios based on what type of ball is transferred. If a green ball is taken from bag \( A \), bag \( B \) contains 5 blue and 10 green balls. If a blue ball is taken, bag \( B \) contains 6 blue and 9 green balls. The probabilities of selecting a green ball from bag \( B \) in each scenario are weighted by the probabilities of transferring each type of ball from bag \( A \). Calculating this gives us an overall probability of approximately \( \frac{62}{100} = \frac{31}{50} \) or 0.62. For part (b): (i) The probability that Team \( M \) wins one game and loses one game can be calculated using the binomial formula. The probability of winning is \( \frac{8}{15} \) and losing is \( \frac{1}{3} \). That means the probability of winning one time (and losing once) in two matches would be the sum of probabilities for the two scenarios (win, lose and lose, win), yielding \( 2 \times \left(\frac{8}{15} \times \frac{1}{3}\right) \) which simplifies to \( \frac{16}{45} \). (ii) The probability that Team \( M \) does not lose both matches is equal to the probability of either winning both matches or winning one and losing one. This can be calculated by determining the probability of winning both matches, which is \( \left(\frac{8}{15}\right)^2 = \frac{64}{225} \), and adding the previously calculated probability of winning one and losing one: \( \frac{64}{225} + \frac{16}{45} = \frac{64}{225} + \frac{80}{225} = \frac{144}{225} \), which reduces to \( \frac{16}{25} \).
