Q.6.13 - A First Course in Probability

Question

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? 

Step-by-Step Solution

Answer

a) probability of home team winning is :$P = 0.8897$

b) P= 0.2818

c) The conditional probability is P = 0.9874

Part (a) - Step 1: To find

The probability that the home team wins.

Part (a) Step 2: Explanation

Let Z be the standard normal random variable:-

Probability of home team Winning :

$X_i$ is the difference of home and visiting team

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

From normal distribution table:

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

Therefore the probability of home team wins is 0.8897.

Part (b) - Step 3: To find

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

Part (b) - Step 4: Explanation

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 ) $=\sum _{i=1}^2{X}_i=-5$

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

Part (c) - Step 5: To find

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.

Part (c) - Step 6: Explanation

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

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

$X_i=5$

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