Question

question five
during a recent basketball season, a player made 206 2 - point shots, 48 3 - point shots and 56 foul shots. total = 310
a) what is the probability the player made a foul shot?
b) what is the probability the player made a 2 - point shot?
c) what is the probability the player made a 2 - point shot or a 3 - point shot?
question six
let p(a)=0.32, p(b)=0.45 and p(a or b)=0.77.
a) are events a and b mutually exclusive?
b) why or why not? explain using probability notation.

Explanation:

Step1: Recall probability formula

The probability formula is $P(E)=\frac{n(E)}{n(S)}$, where $n(E)$ is the number of favorable outcomes and $n(S)$ is the total number of outcomes.

Step2: Calculate probability of foul - shot (Question Five a)

The number of foul - shots $n = 56$, and the total number of shots $n(S)=310$. So $P(\text{foul shot})=\frac{56}{310}=\frac{28}{155}\approx0.181$.

Step3: Calculate probability of 2 - point shot (Question Five b)

The number of 2 - point shots $n = 206$, and $n(S)=310$. So $P(2 - \text{point shot})=\frac{206}{310}=\frac{103}{155}\approx0.665$.

Step4: Calculate probability of 2 - point or 3 - point shot (Question Five c)

The number of 2 - point shots is 206 and the number of 3 - point shots is 48. The number of 2 - point or 3 - point shots $n=206 + 48=254$. So $P(2 - \text{point or }3 - \text{point shot})=\frac{254}{310}=\frac{127}{155}\approx0.819$.

Step5: Check for mutual exclusivity (Question Six a)

Two events $A$ and $B$ are mutually exclusive if $P(A\cup B)=P(A)+P(B)$.

Step6: Calculate $P(A)+P(B)$ (Question Six b)

We know that $P(A) = 0.32$ and $P(B)=0.45$, so $P(A)+P(B)=0.32 + 0.45=0.77$. And $P(A\cup B)=0.77$. Since $P(A\cup B)=P(A)+P(B)$, events $A$ and $B$ are mutually exclusive.

Answer:

Question Five:
a) $\frac{28}{155}\approx0.181$
b) $\frac{103}{155}\approx0.665$
c) $\frac{127}{155}\approx0.819$
Question Six:
a) Yes
b) Since $P(A\cup B)=P(A)+P(B)=0.77$, events $A$ and $B$ are mutually exclusive.