SOLUTION: Find how many 4-digit numbers can be formed from the digits 2,4,6,7 and 9, without repetitions of digits. if one of these numbers are chosen randomly, find the probability that is

Question 352524: Find how many 4-digit numbers can be formed from the digits 2,4,6,7 and 9, without repetitions of digits. if one of these numbers are chosen randomly, find the probability that is
a) greater than 7000
b)an even number
c)greater than 7000 and an even number
then, find the probability that one of the 4-digit numbers formed, chosen at random, has a value greater than 7000 or is an even number

Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
a) The thousands place can be filled in 2 ways (7 or 9). After filling the thousands place, the hundreds, the tens, and the ones places can be filled in 4, 3, and 2 ways, respectively. By the multiplication property of counting, there are

numbers greater than 7000.
b) The ones place can be filled in 3 ways (2, 4, or 6). After filling the ones place, the thousands, hundreds, and the tens places can be filled in 4, 3, and 2 ways, respectively. By the multiplication property of counting, there are

even numbers.
c) The thousands place can be filled in 2 ways (7 or 9). The ones place can be filled in 3 ways (2, 4, or 6). (Note that the two sets of choices are disjoint.) The hundreds place can be filled in 3 ways, the tens place in two ways. By the multiplication property of counting, there are

even numbers that are greater than 7000.
Now from the formula P(A or B) = P(A) + P(B) - P(A and B), we get
P(>7000 or even)= P(>7000)+P(even) - P(>7000 and even).
The number of ways of forming a four-digit number from 5 numbers is
.
Hence P(>7000)=48/120
P(even)=72/120, and P(>7000 and even)=36/120 from
the information in the items above.
Thus, P(>7000 or even)= 48/120 + 72/120 - 36/120 = 84/120 = 7/10.

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