Achiever test 1 - QR #20

[dds10]

---

Members do not see this ad.

Can anyone explain this?

If 1/3 of the cars driven by residents in a specific town are japanese made, what is the probability of seeing at least one japanese car out of every three automobiles on the road?

.
..
...
..
.
Answer : 19/27

 

[americanpierg]

Can anyone explain this?

If 1/3 of the cars driven by residents in a specific town are japanese made, what is the probability of seeing at least one japanese car out of every three automobiles on the road?

.
..
...
..
.
Answer : 19/27

gotta memorize your equations. either way, you can go about this 2 ways.

A) What are the chances of seeing more than 0 (find chances of 0, subtract from 1) <--- easier method
B) What are the chances of seeing 1, seeing 2, or seeing 3 (add them all up)


A)
n!/r!(n-r)! * (probably success)^r * (probability failure)^n-r
3!/0!(3-0)! * (1/3)^0 * (2/3)^3

1-8/27 = 19/27

B)

n!/r!(n-r)! * (probably success)^r * (probability failure)^n-r

3!/1!2! * (1/3)^1 * (2/3)^2 4/9 12/27
3!/2!1! * (1/3)^2 * (2/3)^1 2/9 6/27
3!/3!0! * (1/3)^3 * (2/3)^0 1/27 1/27
19/27

**Notice how they chose nice, small numbers like 1/3 and 3

 

[dds10]

gotta memorize your equations. either way, you can go about this 2 ways.

A) What are the chances of seeing more than 0 (find chances of 0, subtract from 1) <--- easier method
B) What are the chances of seeing 1, seeing 2, or seeing 3 (add them all up)


A)
n!/r!(n-r)! * (probably success)^r * (probability failure)^n-r
3!/0!(3-0)! * (1/3)^0 * (2/3)^3

1-8/27 = 19/27

B)

n!/r!(n-r)! * (probably success)^r * (probability failure)^n-r

3!/1!2! * (1/3)^1 * (2/3)^2 4/9 12/27
3!/2!1! * (1/3)^2 * (2/3)^1 2/9 6/27
3!/3!0! * (1/3)^3 * (2/3)^0 1/27 1/27
19/27

**Notice how they chose nice, small numbers like 1/3 and 3

Thanks! It makes sense, it's just hard for me to remember how to do these.
I always have a hard time with these, I guess i'll look on the net for some explanations.