# which of the following can be the probability of winning the match of a team?

### Mohammed

Guys, does anyone know the answer?

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## 1st Year Probability Question: In a 40 game season, what is the probability of a certain season outcome?

Question There are $40$ games in a season. $30$ are against class A teams which you have a $.4$ probability of winning and $.6$ of losing. $10$ are against class B teams which you have a $.7$

## 1st Year Probability Question: In a 40 game season, what is the probability of a certain season outcome?

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Asked 4 years, 4 months ago

Modified 4 years, 4 months ago

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**Question**

There are 40 40 games in a season. 30 30

are against class A teams which you have a

.4 .4

probability of winning and

.6 .6 of losing. 10 10

are against class B teams which you have a

.7 .7

probability of winning and

.3 .3 of losing.

The season schedule is completely random in regards to what type of team you play.

What is the probability that you:

win 20 20

games against class A teams (and lose

10 10 games) AND win 3 3

games against class B teams (and lose

7 7 games)

**My Attempt**

First I created my own notation...

For example: "A a b b B..." means your first game was against a "class a team" and you won, then you played another "class a" and lost, and then you played 3 "class b teams" and lost, lost, won. I hope that makes sense.

If the probability of winning is always

.5 .5

regardless of the class, then I believe you'd get the answer by dividing the following:

Numerator: 40! 20!10!3!7! 40!20!10!3!7! Denominator: ∑ 10 B=0 ∑ 30 A=0 40! A!(30−A)!B!(10−B)!

∑B=010∑A=03040!A!(30−A)!B!(10−B)!

My thinking here is it's like getting the number of unique permutations of the letters PEPPER.

But since the probability is not

.5 .5

you take the quotient above and multiply it by:

.4 20 .6 10 .7 3 .3 7 .420.610.73.37

Is this the right track? I don't think it is. That "Denominator" above is such a big number I can't do any calculations to do a sanity check. Thanks.

probability Share

asked Dec 22, 2017 at 2:31

HJ_beginner 1,6351 1 gold badge 8 8 silver badges 20 20 bronze badges 2 ( 30 20 ) 0.4 20 0.6 10 ( 10 3 ) 0.7 3 0.3 7

(3020)0.4200.610(103)0.730.37

– Doug M

Dec 22, 2017 at 2:35

Thanks for your help. Does that answer factor in the different schedules you could have? Can you give me the intuition why that doesn't imply your season must be all 30 type a games first and then 10 type b games at the end? That's why I thought you couldn't do that. –

HJ_beginner

Dec 22, 2017 at 2:37

2

The order in which you play the games is completely irrelevant, as long as there are

30 30 type A games and 10 10

type B games, because the win probability for a given game depends only on the type and not on what other games you have played or are going to play. –

Robert Israel

Dec 22, 2017 at 3:01

In the answer Doug M gave the order matters within each type of game doesn't it? For example 30 choose 20? There's only one way to have 30 wins against type A teams (30 choose 30). So is it accurate to say that order matters within type A teams and type B teams but you can mix them freely without worry of order? –

HJ_beginner

Dec 22, 2017 at 6:07

Add a comment

## 1 Answer

1

Between Comments and Answers we have two different "answers" to the Question. Unfortunately, the formal Answer seems to compute to something greater than 1, which can't be the probability of anything. (It has somewhat the form of a multinomial probability, except that the probabilities don't add to 1.)

I believe the Comments are correct, and the numerical answer indicated by @DougM computes to

1.79808× 10 −05 . 1.79808×10−05.

In case more explanation is required, let

A∼Binom(n=30,p=.4) A∼Binom(n=30,p=.4)

denote the number of Class A games won and

B∼Binom(n=10,p=.7) B∼Binom(n=10,p=.7)

the number of Class B games won.

Then we seek P(A=20)×P(B=3)=( 30 20 ) .4 20 .6 10 ×( 10 3 ) .7 3 .3 7

=0.001997491×0.009001692.

P(A=20)×P(B=3)=(3020).420.610×(103).73.37=0.001997491×0.009001692.

In R statistical software:

dbinom(20, 30, .4) * dbinom(3, 10, .7)

## 1.79808e-05

dbinom(20,30,.4);dbinom(3,10,.7)

## 0.001997491 ## 0.009001692 Share

edited Dec 22, 2017 at 6:43

answered Dec 22, 2017 at 5:52

BruceET 49.2k8 8 gold badges 26 26 silver badges 59 59 bronze badges

As you indicated A is a Binomial RV... and order would matter wouldn't it? Because of the 30 choose 20. So order in some sense matters. When you mix A and B, order suddenly doesn't matter? Sorry I'm a novice and I'm trying to build intuition. Thanks. –

HJ_beginner

Dec 22, 2017 at 6:08

In a binomial probability order doesn't matter, only the number of Successes matters. (To the binomial PDF, one may temporarily keep track of order, but then probabilities of outcomes with various orders are merged. For example, with 3 fair coins, P(2 heads) = P(HHT, HTH, THH) = 3(1/8) = 3/8.) –

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## a team has probability 2/3 of winning a game whenever it plays .if the team plays 4 games then the probability that it wins more than half of the games is

Click here👆to get an answer to your question ✍️ a team has probability 2/3 of winning a game whenever it plays .if the team plays 4 games then the probability that it wins more than half of the games is

Question

## a team has probability 2/3 of winning a game whenever it plays .if the team plays 4 games then the probability that it wins more than half of the games is

**A**

## 17/25

**B**

## 15/19

**C**

## 16/27

**D**

## 13/20

Hard Open in App

## If the probability of a team winning is 0.5 and three matches are played, what is the probability that the team will win: exactly one match, at least one match or exactly two matches?

Answer (1 of 3): Let us use binomial distribution. Here. P=q=.5 P(x=1)=3c1(.5)^1(.5)^2 =3×.125 =.375 P(x=2)=3c2(.5)^2(.5)^1 =3×.125 =.375 P(x>=1)=1-p(x=0) 1–3c0(.5)^0(.5)^3 1-.375 =.625

If the probability of a team winning is 0.5 and three matches are played, what is the probability that the team will win: exactly one match, at least one match or exactly two matches?

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3 Answers

Colinjivadi Mahadevan

, former Retired From Life Insurance Corporation of India at Life Insurance Corporation of India (1965-2002)

Answered Nov 20, 2021 · Author has 949 answers and 416.9K answer views

Probability of winning one match=3C1*0.5*0.5*0.5= 0.375

Probability of winning at least one match= 1- Probability of losing all matches=1-3C3*0.5^3= 0.875

Probability of winning exactly two matches=3C2*.5*.5*.5= 0.375

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Partha Chattopadhyay

, studied Mathematics at University of Calcutta

Answered Nov 21, 2021 · Author has 3.5K answers and 4.5M answer views

I. Exactly one match is won

P = 5 C 1 (0.5 ) 1 (0.5 ) 4 =5∗(0.5 ) 5 =5∗0.03125=0.15625

P=5C1(0.5)1(0.5)4=5∗(0.5)5=5∗0.03125=0.15625

II. At least one match is won

Probability of winning at least one match =1- Probability of winning no match =

1−(0.5 ) 5 =0.96875 1−(0.5)5=0.96875

III. Exactly two matches are won

P = 5 C 2 (0.5 ) 2 (0.5 ) 3 =10∗0.03125=0.3125

P=5C2(0.5)2(0.5)3=10∗0.03125=0.3125

Varadarajan Parthasarathy

, B.sc Mathematics, University of Madras (1976)

Answered Nov 20, 2021 · Author has 4.1K answers and 1.3M answer views

Let us use binomial distribution.

Here. P=q=.5

P(x=1)=3c1(.5)^1(.5)^2

=3×.125 =.375

P(x=2)=3c2(.5)^2(.5)^1

=3×.125 =.375 P(x>=1)=1-p(x=0) 1–3c0(.5)^0(.5)^3 1-.375 =.625 Sponsored by USAFIS

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Related answers Related Answer Carter Glass

, studied at Simon Fraser University

Answered 4 years ago · Author has 95 answers and 211.5K answer views

Suppose that A and B are two equally strong teams. Is it more probable that A will beat B in three games out of four or in five games out of seven?

Assuming equal probability of winning, the chance that A wins exactly three of four games is 0.5^4, or 1/16. However, there are four ways they can win three games out of four (4 choose 3), so the chance that A will win 3 out of four games is 4 * 1/16 = 4/16 = 1/4 = 0.25

As for winning exactly five of seven games, we take 0.5^7, which is 1/128. There are 21 different ways for A to do this (7 choose 5), so 21 * 1/128 = 21/128 ~ 0.16

A is more likely to win 3/4 games than to win 5/7, assuming A and B have the same chance of winning.

Related Answer Kevin Spence

, studied at University of Pittsburgh

Answered 1 year ago · Author has 222 answers and 163.8K answer views

Teams A and B are playing a series of games. The series is over as soon as a team wins a total of three games. Team A is two times more likely to win a game than Team B. What is the probability that Team A wins the series?

79% - almost 4 times out of 5.

The key to this is realizing that the number of games will not always be 5.

If A wins in a sweep - you have 2/3*(2/3)*2/3 percent chance of that happening - 8/27, or 29.63%

If A wins in 4, now we have 2/3*(2/3)*2/3*(1/3)*3 - the 1/3 is the chance that B wins a game. Note - there are only 3 ways B can win a game, not 4. B cannot win Game 4 because Game 4 would not be played in case of a sweep. That is why you cannot use a straight Pascal’s triangle to get your coefficients - the 1–4–6–4–1 is not possible if B cannot win Game 4. Anyway, the math is the same as the abo

Related Answer Rohan Das

, studied at Indian Institute of Science, Bangalore

Answered 2 years ago

A certain team wins with probability 0.7, loses with probability 0.2 and ties with probability 0.1. The team plays three games. What is the probability that team wins at least 2 games but not lose?

The team wins at least 2 games but not lose. So there are clearly two cases. First case is, the team wins all three games. Case 2 is, the team wins 2 games and not lose. So the third game should be a draw.

Guys, does anyone know the answer?