Question

A train has 10 cars numbered 1 through 10. If the cars are coupled randomly, what is the probability that the first 3 cars are in car number order?
1/720

2/15

1/5,040

Answer 1

If they are in order the highest number for the first car is 7...and the other cars can only have one specific value because of the first, so

(7/10)(1/9)(1/8)=7/720

Okay, I see that they mean the order has to be 1,2,3 instead of all ordered sets like 7,8,9 and 4,5,6 for example...

So if they have to be 1,2,3 then

(1/10)(1/9)(1/8)=1/720

Answer 2

Answer:

Hence, the probability that the first 3 cars are in car number order is:

$\dfrac{1}{720}$

Step-by-step explanation:

A train has 10 cars numbered 1 through 10.

If the cars are coupled randomly, what is the probability that the first 3 cars are in car number order.

The probability of the three cars is independent and we know that when the events A,B and C are independent then

$P(A\bigcap B\bigcap C)=P(A)\times P(B)\times P(C)$

As, the first car chosen is to be selected among 10 cars.

Hence, the probability is:

$P(A)=\dfrac{1}{10}$

similarly the second car is to be selected among 9 cars.

Hence,

$P(B)=\dfrac{1}{9}$

similarly,

$P(C)=\dfrac{1}{8}$

Hence, the probability that the first 3 cars are in car number order is:

$=\dfrac{1}{10}\times \dfrac{1}{9}\times \dfrac{1}{8}\\\\\\=\dfrac{1}{720}$