Which situation could best be represented by this linear equation? 24x + 38y = 520
2 answers:
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Answer:
1)All possible sequences of events V and M by set played that will result in Player V winning the match = VV; MVV; VMV
2) All possible sequences of events V and M by set played that will result in Player M winning the match:
MM; VMM; MVM
3) The probability player V wins the match = 0.4575
4) Probability that a match between Player V and Player M will consist of 3 sets given that Player V wins the match. = 0.3443
The expected number of sets played when Player V competes in a match with Player M = 2.35
Explanation:
Given:
Event player V wins a set = V
Event player M wins a set = M
1)All possible sequences of events V and M by set played that will result in Player V winning the match = VV; MVV; VMV
2) All possible sequences of events V and M by set played that will result in Player M winning the match:
MM; VMM; MVM
3) The probability player V wins the match:
P(V) = P(VV) + P(MVV) + P(VMV)
P(VV) means probability player V wins first and second set. In the first set she had an equal chance of winning, i.e 50%. We are told if she wins first set, probability of winning second set is 0.60.
Therefore, P(VV) = 0.50 * 0.60
= 0.30
P(MVV) means player V lost the first set, but won the remaining two sets.
P(MVV) = 0.50 *(1-0.70) * 0.45
= 0.0675
VMV means player V won the first set, lost the second, and won the third set.
P(VMV) = 0.50 * (1-0.60) * 0.45
= 0.09
Therefore
P(VV) + P(MVV) + P(VMV)
= 0.30 + 0.0675 + 0.09
= 0.4575
Probability V wins is 0.4575
4) Probability that a match between Player V and Player M will consist of 3 sets given that Player V wins the match.
Here it means player V must lose one of the first two sets, but will still win the match.
Calculating, we have:
= 0.344262 ≈ 0.3443
5) The expected number of sets played when Player V competes in a match with Player M.
Let's first find the probability 2 sets are played.
P(2) = P(VV) + P(MM)
We know P(VV) = 0.30
P(MM) = 0.50 * 0.70 = 0.35
Therefore, P(2) = 0.30 + 0.35
P(2)= 0.65
Let's now find probability of 3 sets happening.
P(3) = P(2) not happening
Therefore
P(3) = 1 - P(2)
P(3) = 1 - 0.65 = 0.35
Since probability of 2 sets is 0.65 and probability of 3 sets is 0.35, the expected number of sets will be :
(2 * 0.65) + (3 * 0.35)
= 1.30 + 1.05 = 2.35
The expected number of sets = 2.35