Write your answer as a whole number and remainder.<br> 11÷6

In the bronze medal game of the 2014 FIBA Asia Cup, and with his team down by two points, Philippines guard Paul Dalistan was fo

dsp73

Complete Question

Question: In the bronze medal game of the 2014 FIBA Asia Cup... In the bronze medal game of the 2014 FIBA Asia Cup, and with his team down by two points, Philippines guard Paul Dalistan was fouled while attempting a three-pointer at the buzzer. He missed the shot and headed to the line for three foul shots.

1. Watch the video of Dalistan preparing to take his first shot. Before he shoots, what is the probability that Philippines will win the game? Lose the game? Send it into overtime? Dalistan’s free throw percentage is 84.6%

Answer:

a

The probability that Philippines will win the game is $P(A) = 0.6055$

b

The probability that Philippines will lose the game is $P(B) = 0.00365$

c

The probability that the game will be sent to overtime is $P(C) = 0.06$

Step-by-step explanation:

From the question we are told that

The probability of Paul Dalistan making a shot is p = 0.846

The probability of Paul Dalistan not making the shot is $q = 1- p = 1- 0.846=0.154$

Paul Dalistan team was down by two points

This mean that for Paul Dalistan's team to win the game he must make the 3 free throws

So the probability that Philippines will win the game is mathematically represented as

$P(A) = p^3$

=> $P(A) = 0.846^3$

=> $P(A) = 0.6055$

Gnerally for Paul Dalistan's team to loss the game it means that Paul Dalistan did not make any of the free shot

Hence the probability that Philippines will loss the game is

$P(B) = q^3$

=> $P(B) = (0.154 )^3$

=> $P(B) = 0.00365$

Generally for the game to be sent to overtime then the game must be tied and for the game to be tied, Paul Dalistan have to make 2 out of the three shots

Hence the probability that game will go to overtime is

$P(C) = 3 * 0.846 * 0.154 * 0.154$

$P(C) = 0.06$

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