A model proposed for NBA basketball supposes that when two teams with roughly the same record play each other, the number of points scored in a quarter by the home team minus the number scored by the visiting team is approximately a normal random variable with mean 1.5 and variance 6. In addition, the model supposes that the point differentials for the four quarters are independent. Assume that this model is correct.

(a) What is the probability that the home team wins?

(b) What is the conditional probability that the home team wins, given that it is behind by 5 points at halftime?

(c) What is the conditional probability that the home team wins, given that it is ahead by 5 points at the end of the first quarter?

The probability that the home team wins.

Let Z be the standard normal random variable:-

Probability of home team Winning :

$X_i$is the difference of home and visiting team

localid="1647576088013" $\begin{array}{r}=P\left\{\frac{\sum _{i=1}^3 X_i-6}{\sqrt{24}}>\frac{-6}{\sqrt{24}}\right\}\\ \approx P(Z>-1.2247)\end{array}$

From normal distribution table:

localid="1647576093710" $\begin{array}{r}P(Z>-1.2287)=1-\mathrm{\Phi }(-1.2247)\\ =\mathrm{\Phi }\left(1.2247\right)\\ =0.8897\end{array}$

Therefore the probability of home team wins is 0.8897.

What is the conditional probability that the home team wins, given that it is behind by 5 points at halftime?

Given that the home team is down by 5 points at halftime

$P\left\{\sum _{i=1}^4{X}_i>0\mid \sum _{i=1}^2{X}_i=-5\right\}$

As after halftime (Two quarters ) $=\sum _{i=1}^2{X}_i=-5$

$\begin{array}{r}p=P\left(X_3+X_4>5\right)\\ =P\left(\frac{X_3+X_4-3}{\sqrt{12}}>\frac{5-3}{\sqrt{12}}\right)\\ =P\left(\frac{X_3-1.5+X_4-1.5}{\sqrt{12}}>\frac{2}{\sqrt{12}}\right)\\ \approx P(Z>0.5774)\\ =1-\mathrm{\Phi }\left(0.5774\right)\\ =1-0.7180\\ p=0.2818\end{array}$

What is the conditional probability that the home team wins, given that it is ahead by 5 points at the end of the first quarter.

$P\left\{\sum _{i=1}^4{X}_i>0\mid X_1=5\right\}=P\left\{X_2+X_3+X_4>-5\right\}$

Therefore $X_1+X_2+X_3+X_4>0$ (for home team win)

$X_i=5$

$\begin{array}{r}\Rightarrow X_2+X_3+X_4>-5\\ p=P\left\{\frac{X_2+X_3+X_4-4.5}{\sqrt{6+6+6}}>\frac{-9.5}{\sqrt{18}}\right\}\\ =P(Z>-2.239)\\ =\mathrm{\Phi }\left(2.239\right)\\ =0.9874\end{array}$