Question

probability (i) Car color preferences change over the years and according to the particular model that the customer selects. In a recent year, 10% of all luxury cars sold were black. If 25 cars of that year and type are randomly selected, find the following probabilities: a. At least five cars are black b. At most six cars are black C. More than four cars are black e 0 d. Exactly four cars are black e. Between three and five cars (inclusive) are black ​

Answer 1

Step-by-step explanation:

Car color preferences change over the years and according to the particular model that the customer selects. In a recent year, 10% of all luxury cars sold were black. If 25 cars of that year and type are randomly selected, find the following probabilities: a. At least five cars are black b. At most six cars are black C. More than four cars are black e 0 d. Exactly four cars are black e. Between three and five cars (inclusive) are black

Answer 2

Given the percentage of black cars sold, find the given probabilities given that $25$ cars are selected at random.

Explanation:

1. here, we can say that the data is in binomial distribution as it has a finite number of sample size 'n' (that is number of cars selected) and equal probability of events of succeeding 'p' (that is the car is black) or failing 'q' (that is the car is not black)
2. here, we have   $n=25,\ \ \ \ \ p=10\%=0.1,\ \ \ \ \ q=1-p=0.9$
3. probability of 'x' number of cars to to be black is given by ,                                [tex]P(x)= {^nC_x}p^xq^{n-x}\\ P(x)= {^{25}C_x}(0.1)^x(0.9)^{25-x}[/tex]                       $(here,\ ^nC_r=\frac{n!}{r!(n-r)!} )$        
4. hence, probability that at least five cars are black is given by                                  [tex]P(x\geq 5)=1-P(x<5)\\ P(x\geq5)= 1-[P(1)+P(2)+P(3)+P(4)]\\ P(x\geq 5)= 1-[ {^{25}C_0}(0.1)^0(0.9)^{25}+\ {^{25}C_1}(0.1)^1(0.9)^{24}+\ ^{25}C_2}(0.1)^2(0.9)^{23}+\ ^{25}C_3}(0.1)^3(0.9)^{22}+ \ ^{25}C_4}(0.1)^4(0.9)^{21}]\\\\ P(x\geq 5)= 1-[0.902] \\ P(x\geq 5)=0.098[/tex]  
5. now probability of at most six cars to be black is given by,                     $P(x\leq 6)=P(1)+P(2)+P(3)+P(4)+P(5)+P(6)\\\\P(x\leq 6)= {^{25}C_0}(0.1)^0(0.9)^{25}\ +{^{25}C_1}(0.1)^1(0.9)^{24}\ +^{25}C_2}(0.1)^2(0.9)^{23}\ +\ ^{25}C_3}(0.1)^3(0.9)^{22}\ +^{25}C_4}(0.1)^4(0.9)^{21}\ +^{25}C_5}(0.1)^5(0.9)^{20}\ +^{25}C_6}(0.1)^6(0.9)^{19}\\\\P(x\leq 6)= 0.99$  
6. now probability that more than four cars are black is given by,                         [tex]P(x> 4)=1-P(x\leq 4)\\P(x>4)= 1-P(x<5)\\ P(x>4)=0.098[/tex]          
7. now probability that exactly four cars are black is given by,                                   [tex]P(x=4)={^{25}C_4(0.1)^4(0.9)^{21} \\ P(x=4)= 0.138[/tex]
8. now probability that between three to five cars are black is,                            [tex]P(3\leq x\leq 5)=P(3)+P(4)+P(5)\\ P(3\leq x\leq 5)= ^{25}C_3}(0.1)^3(0.9)^{22}+ \ ^{25}C_4}(0.1)^4(0.9)^{21}+\ ^{25}C_5(0.1)^5(0.9)^{20}\\\\P(3 \leq x\leq 5)= 0.429[/tex]  
9. The final probabilities are rounded to three decimal places:                                 (a)->$0.098$ (b)->$0.99$ (c)->$0.098$ (d)->$0.138$ (e)->$0.429$