1. ## Probability

A four digit number is to be made by choosing four different digits from the set {1,2,3,4,5,6,7}. Find the probability that the no. is even and greater than 6000.

2. Hello, BabyMilo

A 4-digit number is to be made by choosing four different digits from the set {1,2,3,4,5,6,7}.
Find the probability that the number is even and greater than 6000.
There are: . $_7P_4 \:=\:\frac{7!}{3!} \:=\:840$ possible 4-digit numbers.

How many of them are even and greater than 6000?

There are two cases to consider.

[1] The number begins with "6": .6 _ _ _

We have: .{1, 2, 3, 4, 5, 7} to work with.

The last digit must be even: 2 choices.

The middle two digits are selected from the remaining 5 digits: . $_5P_2= 20$ choices.

Hence, there are: . $2\cdot20 = 40$ even numbers that begin with "6".

[2] The number begins with "7": .7 _ _ _

We have: .{1, 2, 3, 4, 5, 6} to work with.

The last digit must be even: 3 choices.

The middle two digits are selected from the remaining 5 digits: . $_5P_2 = 20$ choices.

Hence, there are: . $3\cdot20 = 60$ even numbers that begin with "7".

Then there are: . $40 + 60 \:=\:100$ even numbers greater than 6000.

Therefore: . $P\bigg[(\text{even}) \wedge (> 6000)\bigg] \;=\;\frac{100}{840} \:=\:\frac{5}{42}$