# StatisticsNBA playoffs are at full swing and Kobe Bryant (Lakers) shoots 83.8 % from the free line throw. Down two points with only seconds with only seconds left in their first round game against...

Statistics

NBA playoffs are at full swing and Kobe Bryant (Lakers) shoots 83.8 % from the free line throw. Down two points with only seconds with only seconds left in their first round game against Denver, Kobe is fouled while attempting a buzzer beating 3-pointer. He stands at the foul line ready to shoot three shots each worth a single point.

a)Make a probability tree and determine the probability that he makes 0,1,2,3 of the free throws.

b) What is the probability that the Lakers lose the game in regulation (meaning before overtime)?

c) If the games goes overtime, there is a 86.3% chance that the Lakers will win. What is the probability that the game will be won by them?

Asked on by juno60

### 1 Answer |Add Yours

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

(1) Create a tree diagram to determine the probabilities of hitting 0,1,2,or 3 shots:

-.838-H--.838-H--.838-H  =======> P(HHH)`=(.838)^3~~.5885`
|
--.062-M ========>P(HHM)`~~.1138`
|
--.162-M--.838-H  ========>P(HMH)`~~.1138`
|
--.162-M ========>P(HMM)`~~.0220`

-.062-M--.838-H--.838-H========>P(MHH)`~~.1138`
|
--.162-M=========>P(MHM)`~~.0220`
|
--.162-M--.838-H=========>P(MMH)`~~.0220`
|
--.162-M=========>P(MMM)`~~.0043`

So the probability of 0 baskets is approximately .0043
the probability of 1 basket is approx 3(.0220)=.0660
the probability of 2 baskets is approx 3(.1138)=.3414
the probability of 3 baskets is approximately .5885

Note that each branch is "weighted" with the probability that it occurs.

(2) The Lakers lose the game in regulation if he misses two or more shots (or makes 1 or fewer shots).

Then the probability of losing is approx. .066+.0043=.0703

The events are independent (he cannot make exactly 1 shot and exactly 0 shots) so you add the probabilities.

(3) The probability of the Lakers winning is the sum of the probability of winning in regulation and the probability of winning in overtime (if it goes to overtime)

P(W)=P(WR)+P(WO)
=.5885 + .3414(.863)
=.8831

So the probability of a win is approximately 88.3%

The probability of winning in regulation is straightforward; the probability of winning in overtime depends on getting to overtime in the first place, thus its probability is weighted.

Sources:

We’ve answered 317,520 questions. We can answer yours, too.