Respuesta :
Answer:
- (a) 0.42
- (b) 0.028
- (c) 4
Explanation:
Since each throw is independent of the last throw and there are only two options, make and no make it, you can use the equations for binomial experiments.
Call X the random variable, making a throw, p the probability of making it and (1-p) the probability of not making it, then the probability of making exacty x throws in n attempts is:
[tex]P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^n[/tex]
[tex]n=7\\\\p=0.6\\\\1-p=0.4[/tex]
[tex]C(n.x)=\dfrac{n!}{x!\cdot (n-x)!}[/tex]
a. What is the probability that he will make more than four of his free throws?
The probability of making more than four of the free throws is the sum of the probabilities of making exactly five, exactly six, and exactly seven throws.
[tex]P(X>4)=P(X=5)+P(X=6)+P(X=7)[/tex]
[tex]P(X=5)=\dfrac{7!}{5!\cdot (7-5)!}\times (0.6)^5\times (0.4)^2=0.2612736[/tex]
[tex]P(X=6)=\dfrac{7!}{6!\cdot (7-6)!}\times (0.6)^6\times (0.4)^1=0.1306368[/tex]
[tex]P(X=7)=\dfrac{7!}{7!\cdot (7-7)!}\times (0.6)^7\times (0.4)^0=0.0279936[/tex]
[tex]P(X>4)=0.2612736+0.1306368+0.0279936=0.4199\approx 0.42[/tex]
b. What is the probability that he will make all of his free throws?
This was calculated above. It is P(X=7)
- P(X=7) = 0.02799936 ≈ 0.028
c. How many free throws should we expect Thad to make of the seven attempts during this game?
The expected number of free throws is the sum of the products of each probability by the number of succesful throws:
[tex]E(X)=\sum\limits^7_0 {x}\cdot p(x)[/tex]
Then, you need P(X=0), P(X=1), P(X=2), P(X=3), P(X=4), P(X=5), PX=(X=6), and P(X=7).
Using the same method that in the part (a), we can find):
- P(X=0) =0.0016384
- P(X=1) = 0.0172032
- P(X=2) = 0.0774144
- P(X=3) = 0.193536
- P(X=4) = 0.290304
From above:
- P(X=5) = 0.2612736
- P(X=6) = 0.1306368
- P(X=7) = 0.0279936
Now substitute in the formula for E(X)
[tex]E(X)=0\times 0.0016+1\times 0.017+2\times 0.077+3\times 0.19+...[/tex]
(I do not show the entire numbers due to space reasons)
The sum is 4.2
Hence, we should expect Thad to make 4 throws of the seven attemts during the game.
a. The probability for more than four of his free throws would be 0.4199
b. The probability for all free throws would be 0.02799
c. The free throws after seven attempts would be 4.2
What is the probability that he will make more than four of his free throws?
The value of p is given as 0.6 and the value n is 7.
[tex]p(x)=\frac{n!}{(n-x)! x!} (p)^{x} (1-p)^{n-x} \\\frac{7!}{5! 2!} (0.6)^{7} (0.4)^{0} \\p=0.4199[/tex]
What is the probability that he will make all of his free throws?
[tex]p=(o.6)^{7} \\=0.02799[/tex]
How many free throws should we expect Thad to make of the seven attempts during this game?
The expected value would be computed as:
[tex]n*p\\=7*0.6\\=4.2[/tex]
Learn more calculations of probability here:
https://brainly.com/question/13018489
