Please answer problem parts a-i, b, and c! Thank you! Two horses, named Gibbs an
ID: 3201517 • Letter: P
Question
Please answer problem parts a-i, b, and c! Thank you!
Two horses, named Gibbs and Boltz, have run 12 head-to-head races in the past. Gibbs has won 5 of these. However, he appears to do better against Boltz when the track is wet (i.e., when it has rained just prior to the race). Table 1 shows the data that correlates race wins with track conditions (R=rain before the race, S=sun before the race). a) From the above table, calculate: i) The joint probability p_W + R that it rains and Gibbs wins. ii) The probability that Gibbs wins (independent of track conditions), P_W. iii) The probability that it rains (independent of which horse wins), P_R. iv) The probability that Gibbs wins when it rains, p(W|R). v) The probability that it rains when Gibbs wins, p(R|W). b) Using the numerical results obtained in part a), confirm that p_W + R = p_(W|R) P_R = p_(R|W) P_W c) If the track is wet the day of the next race (i.e., it rained the night before), what odds (i.e., probability) does Gibbs have of winning the race?Explanation / Answer
a)
i) The joint probability that it rains and Gibbs wins : pW+R = no of times Gibbs wins when it rains / total number of races
No of times Gibbs wins when it rains = 3 times (1st race, 4th race, 12th race)
Total number of races = 12
pW+R = 3 wins / 12 races = 1/4
b) p(W|R) = Probability that Gibbs wins when it rains = 3/4
PR = Probability that it rains = 4/12 = 1/3
p(R|W) = Probability that it rains when Gibbs wins = 3/5
PW = Probability that Gibbs wins = 5/12
p(W|R) PR = 3/4 * 1/3 = 1/4
p(R|W) PW= 3/5 * 5/12 = 1/4
Therefore pW+R = p(W|R) PR = p(R|W) PW
c) If the track is wet the next day, odds of Gibbs winning the race= p(W|R) = 3/4