Q. 5.1 - A First Course in Probability

Question

The number of minutes of playing time of a certain high school basketball player in a randomly chosen game is a random variable whose probability density function is given in the following figure: 

                     

Find the probability that the player plays 

(a) more than $15$ minutes;

(b) between $20 and 35$ minutes;

(c) less than $30$ minutes;

(d) more than $36$ minutes 

Step-by-Step Solution

Answer

(a) The probability that the player plays more than $15$ minutes is $0.875$ 

(b) The probability that the player plays between $20 and 35$ is $0.625$

(c) The probability that the player plays less than $30$minutes is $0.75$

(d) The probability that the player plays more than $36$ minutes is $0.1$

Step1: Find the probability that the player plays more than 15 minutes (part a)

Formalize the given probability function. As may be observed from the graph,

$f\left(x\right)=0.025, x\in [10,20)\cup (30,40]$

$f\left(x\right)=0.05, x\in [20,30]$

$f\left(x\right)=0$, otherwise

$X$ is the random variable with the density function defined $P$. The needed probabilities are calculated as the integrals of the density function $f$ over the relevant intervals.

$$\begin{gathered} P(X>15)=1-P(X\le 15) \\ =1-{\int }_{10}^{15}f\left(x\right)dx \\ =1-{\int }_{10}^{15}0.025dx \end{gathered}$$

$$\begin{gathered} =1-0.025\times 5 \\ =0.875 \end{gathered}$$

Step2: Find the probability that the player plays between 20   a n d   35 minutes (part b)

$P(X\in (20,35\left)\right)=P(X\in (20,30\left)\right)+P(X\in (30,35\left)\right)$

$={\int }_{20}^{30}f\left(x\right)dx+{\int }_{30}^{35}f\left(x\right)dx$

$$\begin{gathered} =0.05\times 10+0.025\times 5 \\ =0.625 \end{gathered}$$

Step3: Find the probability that the player plays less than 30 minutes (part c)

$$\begin{gathered} P(X<30)=1-P(X\ge 30) \\ =1-{\int }_{30}^{40}f\left(x\right)dx \\ =1-{\int }_{30}^{40}0.025dx \end{gathered}$$

$$\begin{gathered} =1-0.025\times 10 \\ =0.75 \end{gathered}$$

step4: Find the probability that the player plays more than 36 minutes (part d)

$$\begin{gathered} P(X>36)={\int }_{36}^{40}f\left(x\right)dx \\ ={\int }_{36}^{40}0.025dx \\ =0.025\times 4 \\ =0.1 \end{gathered}$$