Respuesta :
Answer:
0.8343
Step-by-step explanation:
From the question, we have the following values:
Probability of vehicles that pass within the check point that are from within the state = 75% = 0.75
Probability of vehicles that pass within the check point that are from outsode the state = 100 - 75 = 25% = 0.25
P = 0.25
n = number of random variables = 9
The probability that fewer than 4 of the next 9 vehicles are from out of state is calculated as:
P < 4 = P ≤ 3
n = 9
P(x) = n!/(n - x)! x! × p^x × q^(n - x)
x = 3
p = 0.25
q = 0.75
P(x) = 9! /(9 - 3)! × 3! × 0.25^3 × 0.75^(9 - 3)
P(x) =0.8343
The probability that fewer than 4 (x<4) of the next 9 vehicles are from out of state is 0.83427.
Given information:
75% of the vehicles passing through a checkpoint are from within the state.
So, the probability that the vehicle is from within the state is 0.75.
The probability that the vehicle is from outside the state will be 1-0.75=0.25.
Now, let x be the random variable. So, the value of n=9. and x<4
It is required to calculate the probability that fewer than 4 of the next 9 vehicles are from out of state.
So, [tex]x< 4[/tex], p=0.25 and q=0.75.
So, the required probability can be calculated as,
[tex]P(x\le3) =\sum ^nC_x\times p^x \times q^{(n - x)}\\P(x\le3)=\sum\dfrac{n!}{(n - x)! x!} \times p^x \times q^{(n - x)}\\P(x\le3)= \dfrac{9!}{(9 - 3)! 3!} \times 0.25^3 \times 0.75^{(9 - 3)}+\dfrac{9!}{(9 - 2)! 2!} \times 0.25^2 \times 0.75^{(9 - 2)}+\dfrac{9!}{(9 - 1)! 1!} \times 0.25^1 \times 0.75^{(9 - 1)}+\dfrac{9!}{(9 - 0)! 0!} \times 0.25^0 \times 0.75^{(9 - 0)}\\P(x\le3)=0.83427[/tex]
Therefore, the probability that fewer than 4 of the next 9 vehicles are from out of state is 0.83427.
For more details, refer to the link:
https://brainly.com/question/14282621
