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A team $T$ has probability $0.3$ of winning a football game. Each season comprises $28$ games.

1) What is the probability of the team $T$ winning $6$ games in the first half of the season.

2) How to determine the probability that the team T wins an average of $12$ games per season over the next three seasons ?

My try:

1) Let $X$ be the number of won games, then $X$ has a Binomial distribution $B(14;0.3)$ so $p(X=6)={14\choose 6} (0.3)^6*(0.7)^8$

2) For each season $i=1..3$, let $X_i\sim B(28;0.3)$ the binomial distribution counting number of won games. Then the mean of the $X_i$'s denoted $\bar X$ is a random variable having a $t-$ distribution with mean $28*0.3$ and variance $(28*0.3*0.7)/3$ now the question is $p(\bar X=12)$ which can be solved looking to the $t$ scores on the table. is this correct ? Thank you for your help!

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Your first part is absolutely correct.

But for second part you can use binomial distribution again because question is asking for average of 12 games to be won.

Total no. of games in 3 seasons $=28*3=84$

Total no. of games to be won $=12*3=36$

So P(average of 12 games won in three season) $\ =\ ^{84}C_{36}(0.3)^{36}(0.7)^{48}$

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Using the t-distribution is not a good idea. There is no reason why the sum of three binomial distributed random variables could be approximated by the t-distribution.

It is better to notice that $Y=X_1+X_2+X_3$ is distributed as $Y\sim Bin(84,0.3)$

On average 12 games per season has to be won. That means that the sum of won games is 36.

Therefore the probability to win 12 games on average per season is

$$P(Y=36)=\binom{84}{36}\cdot 0.3^{36}\cdot 0.7^{84-36}\approx0.395\%$$

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    $\begingroup$ it is not the sum which is distributed as a $t$-distribution but the mean $\frac{X_1+X_2+X_3}{3}$. We know the mean of any distribution has a normal distribution when the sample is large but here the sample is small ( 3 seasons) that's why i took a $t$-distribution. $\endgroup$
    – palio
    Apr 21 '17 at 20:00
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    $\begingroup$ I understood your idea. If you have sufficiently many random variables then the mean and the sum are normally distributed, but with different parameters. And you are right that three random variables are not enough. $\endgroup$ Apr 21 '17 at 20:07

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