Question

Kevin, LeBron and Stephen take turns shooting a basketball (in that order) and have probabilities of 0.55, 0.65 and 0.7, respectively, of scoring a basket. Compute the probabilities for each player to get the first basket.

Answer 1

The order of shooting a basketball is Kevin, LeBron and Stephen. The probability that Kevin scores a basket is 0.55, that is, probability that Kevin does not score a basket is (1 -0.55 = 0.45). The probability that LeBron scores a basket is 0.65, that is, probability that LeBron does not score a basket is (1 -0.65 = 0.35). The probability that Stephen scores a basket is 0.7, that is, probability that Stephen does not score a basket is (1 -0.7 = 0.3). Now, the probability that Kevin gets the first basket is obtained as follows: 0.55+(1-0.55)(1-0.65)(1-0.7)0.55 +(1-0.55)' (1-0.65)*(1–0.7)*0.55... =0.55][(1–0.55)(1-0.65)(1-0.7)] =0.550 (Sum of infinite GP series) (1-(1-0.55)(1-0.65)(1–0.7)] 0.55 *1-0.45x0.35x0.3 0.55 0.95275 = 0.5772 Hence, the probability that Kevin gets the first basket is 0.5772. Now, the probability that LeBron gets the first basket is obtained as follows: (1–0.55)0.65+ (1-0.55)(1-0.65)(1–0.7)](1–0.55)0.65+... = (1–0.55)0.65 € [(1–0.55)(1-0.65)(1-0.7)]" =(1-0.550.65 (Sum of infinite GP series) (1-(1-0.55)(1-0.65)(1-0.7)] 0.2925 0.95275 = 0.30700 Hence, the probability that LeBron gets the first basket is 0.3070. Now, the probability that Stephen gets the first basket is obtained as follows: (1–0.55)(1–0.65)0.7+[(1–0.55)(1–0.65)(1–0.7)](1–0.55)(1–0.65)0.75 +... =(1-0.55)(1-0.55)(1-0.65)0.7][(1–0.55)(1–0.65)(1–0.7)] = (1-0.55)(1-0.55)(1-0.65)0.7 (Sum of infinite GP series) "[1-(1-0.55)(1–0.65)(1–0.7)] 0.0496 0.95275 = 0.05205 Hence, the probability that Stephen gets the first basket is 0.05205.