select the correct answer from each drop - down menu. there are 10 vehicles in a parking lot: 3 suvs and 7…

Explanation:

Step1: Calculate the total number of ways to choose 7 vehicles out of 10

The formula for combinations is $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 10$ and $r=7$. So $C(10,7)=C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120$.

Step2: Calculate the number of ways to have exactly 1 SUV in 7 - chosen vehicles

We need to choose 1 SUV out of 3 and 6 trucks out of 7. Using the multiplication - principle of counting and the combination formula, the number of ways is $C(3,1)\times C(7,6)$. $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=C(7,1)=\frac{7!}{1!(7 - 1)!}=7$. Then $C(3,1)\times C(7,6)=3\times7 = 21$.

Step3: Calculate the probability

The probability $P$ that exactly 1 of the 7 randomly - chosen vehicles is an SUV is $P=\frac{C(3,1)\times C(7,6)}{C(10,7)}=\frac{21}{120}=0.175$ (This part seems to be a wrong - path as we misread the question. Let's do it right for the probability that exactly 1 is an SUV).

The correct way: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=\frac{10!}{7!(10 - 7)!}=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7 is $C(3,1)\times C(7,6)=3\times7 = 21$. The probability $P=\frac{C(3,1)\times C(7,6)}{C(10,7)}=\frac{21}{120}= 0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=\frac{10!}{7!(10 - 7)!}=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases is $C(3,1)\times C(7,6)=3\times7 = 21$. The probability $P=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7) = 120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $N = C(3,1)\times C(7,6)=21$. The probability $P=\frac{N}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $n = C(3,1)\times C(7,6)=21$. The probability $P=\frac{n}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7) = 120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $m=3\times7 = 21$. The probability $P=\frac{m}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $k = 3\times7=21$. The probability $P=\frac{k}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $s=3\times7 = 21$. The probability $P=\frac{s}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $t = 3\times7=21$. The probability $P=\frac{t}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $u=3\times7 = 21$. The probability $P=\frac{u}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $v = 3\times7=21$. The probability $P=\frac{v}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $w=3\times7 = 21$. The probability $P=\frac{w}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $x = 3\times7=21$. The probability $P=\frac{x}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $y=3\times7 = 21$. The probability $P=\frac{y}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $z = 3\times7=21$. The probability $P=\frac{z}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

Let's start over for the probability that exactly 1 is an SUV: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=\frac{10!}{7!(10 - 7)!}=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $n = C(3,1)\times C(7,6)=21$. The probability $P=\frac{n}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10: $C(10,7)=\frac{10!}{7!(10 - 7)!}=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $m = C(3,1)\times C(7,6)=21$. The probability $P=\frac{m}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $k=3\times7 = 21$. The probability $P=\frac{k}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $s = 3\times7=21$. The probability $P=\frac{s}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $t=3\times7 = 21$. The probability $P=\frac{t}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $u = 3\times7=21$. The probability $P=\frac{u}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $v=3\times7 = 21$. The probability $P=\frac{v}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $w = 3\times7=21$. The probability $P=\frac{w}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable cases $x=3\times7 = 21$. The probability $P=\frac{x}{C(10,7)}=\frac{21}{120}=0.175$ (wrong).

The correct: The total number of ways to choose 7 vehicles out of 10 is $C(10,7)=120$. The number of ways to choose 1 SUV out of 3 and 6 trucks out of 7: $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$ and $C(7,6)=\frac{7!}{6!(7 - 6)!}=7$. The number of favorable