# Thread: Probability- Explained?

1. ## Probability- Explained?

I was shown how to do this question but I am not sure why it was done like this.

Four numbers are drawn at random from a box of ten numbers 0,1,...,9.

Find the probability that the largest number drawn is a six with replacement.

S={0,1,2,3,4,5,6,7,8,9} #S=10

A:Set of ordered 4-tuples out of S_1={0,1,2,3,4,5,6} #S_1=7

B:Set of ordered 4 tuples out of S_2={0,1,2,3,4,5} #S_2=6

Then,
P(largest of four numbers drawn with replacement is 6)=
[#(A)-#(B)]/10^4, =[(7^4)-(6^4)]/10^4

What I don't understand is why is it [#(A)-#(B)].

2. I would treat this as a multinomial with 3 groups.
I would add the probability of exactly 1, exactly 2, exactly 3 or exactly 4 times we select a 6 as the highest number.

${4 \choose 3,1,0}(.6)^3(.1)^1(.3)^0+{4 \choose 2,2,0}(.6)^2(.1)^2(.3)^0$

$+{4 \choose 1,3,0}(.6)^1(.1)^3(.3)^0+{4\choose 0,4,0}(.6)^0(.1)^4(.3)^0$

$=.0864+.0216+.0024+.0001=.1105$

The $[(7^4)-(6^4)]/10^4$ is the different of the two sets.
If you list all possible sets of four numbers and you first select any number from 0 to 6, but then you remove
all of those that only consist of 0 through 5 you end up with at least one 6 somewhere.

3. Originally Posted by matheagle
I would treat this as a multinomial with 3 groups.
I would add the probability of exactly 1, exactly 2, exactly 3 or exactly 4 times we select a 6 as the highest number.

${4 \choose 3,1,0}(.6)^3(.1)^1(.3)^0+{4 \choose 2,2,0}(.6)^2(.1)^2(.3)^0$

$+{4 \choose 1,3,0}(.6)^1(.1)^3(.3)^0+{4\choose 0,4,0}(.6)^0(.1)^4(.3)^0$

$=.0864+.0216+.0024+.0001=.1105$

The $[(7^4)-(6^4)]/10^4$ in the different is the two sets.
If you list all possible sets of four numbers and you first select any number from 0 to 6, but then you remove
all of those that only consist of 0 through 5 you end up with at least one 6 somewhere.
Thanks so much, I really appreciate this =)