91Ó°ÊÓ

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" $\begin{array}{r}=P\left\{\frac{\underset{i=1}{\overset{3}{∑}} X_i−6}{\sqrt{24}}>\frac{−6}{\sqrt{24}}\right\}\\ ≈P(Z>−1.2247)\end{array}$

From normal distribution table:

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

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

$P\left\{\underset{i=1}{\overset{4}{∑}}{X}_i>0âˆ\pounds \underset{i=1}{\overset{2}{∑}}{X}_i=-5\right\}$

As after halftime (Two quarters ) $=\underset{i=1}{\overset{2}{∑}}{X}_i=-5$

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

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.

$P\left\{\underset{i=1}{\overset{4}{∑}}{X}_i>0âˆ\pounds X_1=5\right\}=P\left\{X_2+X_3+X_4>-5\right\}$

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

$X_i=5$

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