a) The probability that Kim wins the match is 0.5747
b)The probability that Kim wins the match in exactly 2 sets is 0.3025
c) The probability that 3 sets are played is 0.495
Step-by-step explanation:
P(Kim Wins)=0.55
P(Susan Wins)=1-0.55=0.45
The possible outcomes are:
KSK,KSS,SKK,SKS,KK,SS
(a) Kim's wins:
P(KK)+P(KSK)+P(SKK)
P(Kim wins first two sets) P(KK)=0.55×0.55=0.3025
P(Kim wins first set, Susan wins second set, Kim wins third set)
P(KSK)=0.55×0.45×0.55≈0.1361
P(Susan wins first set, Kim wins second and third set)
P(SKK)= 0.45×0.55×0.55≈0.1361
Therefore:
P(Kim wins)
= 0.3025+0.1361+0.1361=0.5747
b) Kim wins the match in exactly 2 sets
P(Kim wins first two sets) P(KK)=0.55×0.55=0.3025
(c) 3 sets are played
For three sets to be played, the possible events are:
KSK,KSS,SKK,SKS
P(Kim wins first set, Susan wins second set, Kim wins third set)=
P(KSK)=0.55×0.45×0.55≈0.1361
P(Susan wins first set, Kim wins second and third set)= P(SKK)=0.45X0.55X0.55=0.1361
P(Kim wins first set, Susan wins second and third set)=
P(KSS)=0.55×0.45×0.45≈0.1114
P(Susan wins first set, Kim wins second set and Susan wins third set)=
P(SKS)=0.45×0.55×0.45≈0.1114
Therefore:
P(3 sets are played)
=0.1361+0.1361+0.1114+0.1114
=0.495
Answer:
a)2.496 b) 1.37225
Step-by-step explanation:
The complete question is:
Kim and Susan are playing a tennis match where the winner must win 2 sets in order to win the match. For each set, the probability that Kim wins the set is 0.55. The probability of Kim winning the set is not affected by who has won any previous sets.
a) calculate the expected value of the number of sets played
b) calculate the expected value of the number of sets that Kim wins
K: probabiltiy that Kim wins=0.55
S: probabiltiy that Susan wins=0.45
See the probability tree diagram attached for the sequence of events
a) Let X be the number of matches played
X=2,3
P(X=2)= P(Kand K) + P(S and S)
= 0.55× 0.55 + 0.45× 0.45
= 0.505
P(X=3)= P(K and S and K) + P(K and S and S) +P(S and K and K) + P(S and K and S)
= 0.55× 0.45× 0.55 + 0.55 ×0.45 ×0.45 + 0.45× 0.55× 0.55 + 0.45× 0.55× 0.45
= 0.495
E(X) =P(X=2) ×2 + P(X=3) × 3
= 0.505 ×2 + 0.495 × 3
= 2.495
b) X: number of sets Kim wins
X=1,2
P(X=1) = P(K and S and S) + P(S and K and S)
= 0.55× 0.45× 0.45 + 0.45× 0.55× 0.45
= 0.22275
P(X=2) = P(K and K ) + P(K and S and K) + P(S and K and K )
= 0.55 ×0.55 + 0.55× 0.45× 0.55 + 0.45× 0.55× 0.55
= 0.57475
E(X)= P(X=1) ×1 + P(X=2) × 2
= 1.37225
