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Player V and Player M have competed against each other many times. Historical data show that each player is equally likely to win the first set. If Player V wins the first set, the probability that she will win the second set is 0.60. If Player V loses the first set, the probability that she will lose the second set is 0.70. If Player V wins exactly one of the first two sets, the probability that she will win the third set is 0.45

What is the probability that Player V will win a match against Player M?

Respuesta :

skyluke89

Answer:

0.46 (46%)

Step-by-step explanation:

We have the following data:

- Probability that player V wins the first set:

[tex]p(1)=0.50[/tex]

(because the text says the two players are equally likely to win the first set)

- Probability that player V wins the 2nd set if he has won the 1st set:

[tex]p(2|1)=0.60[/tex]

So, the probability that player V wins the first 2 sets is:

[tex]p(12)=p(1)\cdot p(2|1)=(0.50)(0.60)=0.30[/tex] (1)

Instead, the probability that player V loses the 2nd set if he has won the 1st set is 0.40 (=1-0.60), so

[tex]p(2^c|1)=0.40[/tex]

So, the probabiity that player V winse the 1st set but loses the 2nd set is

[tex]p(12^c)=p(1)\cdot p(2^c|1)=(0.50)(0.40)=0.20[/tex] (2)

Also, we have:

- Probability that player V loses the 1st set:

[tex]p(1^c)=0.50[/tex]

- Probability that she will lose the 2nd set in this case is 0.70, it means that the probability that she will win the 2nd set if she lost the 1st set is 0.30, so:

[tex]p(2|1^c)=0.30[/tex]

So, the probability that she will lose the 1st set and win the 2nd set is:

[tex]p(1^c2)=p(1^c)\cdot p(2|1^c)=(0.50)(0.30)=0.15[/tex] (3)

Combined together (2) and (3), this means that the probability that player V wins exactly 1 set out of the first two sets is:

[tex]p(1/2)=p(12^c)+p(1^c2)=0.20+0.15=0.35[/tex] (4)

At this point, the probability that she will win the 3rd set is

[tex]p(3)=0.45[/tex]

This means that the overall probability that she will win the 3rd set if she won exacty 1 of the first 2 sets is:

[tex]p(1/2,3) = p(1/2)\cdot p(3)=(0.35)(0.45)=0.16[/tex] (5)

So, the overall probability that player V will win a match against player M is the sum of (1) and (5):

[tex]p(W)=p(12)+p(1/2,3)=0.30+0.16=0.46[/tex]

The probability that Player V will win a match against Player M according to the historical data is 0.4575.

What do you mean by the probability of an event?

Probability of an event is the ratio of number of favourable or required outcome to the total number of outcome can be happen in that event.

Player V and Player M have competed against each other many times.

Historical data show that each player is equally likely to win the first set. Thus, player we and player 2 both has 1/2 probability to win the first set.

[tex]P(V1)=\dfrac{1}{2}\\P(M1)=\dfrac{1}{2}[/tex]

  • If Player V wins the first set, the probability that she will win the second set is 0.60.

Thus, the probability of V wins both set and win the game is,

[tex]P(VV)=\dfrac{1}{2}\times0.6\\P(VV)=0.3[/tex]

  • If Player V loses the first set, the probability that she will lose the second set is 0.70.
  • If Player V wins exactly one of the first two sets, the probability that she will win the third set is 0.45

Thus, the probability that V lose first set but win the second and third set to win the game is,

[tex]P(MVV)=0.50\times(1-0.70)\times0.45\\P(MVV)=0.0675[/tex]

The probability that V win first set, lose second set and again win last set to win the game is,

[tex]P(MVV)=0.50\times(1-0.60)\times0.45\\P(MVV)=0.09[/tex]

The probability that V win match is,

[tex]P(V)=P(VV)+P(MVV)+P(VMV)\\P(V)=0.30+0.0675+0.09\\P(V)=0.4575[/tex]

Thus, the probability that Player V will win a match against Player M according to the historical data is 0.4575.

Learn more about the probability here;

https://brainly.com/question/24756209

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