Question

how many different numbers can be formed by taking 1, 2, 3 and 4 digits from 1,2,7,8. If repetions are not allowed. Find the probabaility that a number chosen at randon is greater than 200.

Answer 1

Do each case separately, to demonstrate the method I will solve it for 2 digits.

So we want to know how many 2-digit numbers we can create by choosing from the numbers 1, 2, 7, 8. Well, the first digit can be any of them, so we have 4 choices for the first digit. For the second digit we now only have 3 numbers left to pick from. Since we can combine any of the 4 first digits with any of the 3 second digits we get 3*4=12 total numbers.

To figure out the probability, just figure out how many of the combinations are greater than 200.

Answer 2

1 didgit numbers are 4P1 = 4!/(4-1)!=4

2 digit = 4P2 = 4!/2!= 12

3 digit = 4P3 = 4!/1! = 24

4 digit = 4! = 24

total will be = 64 i hope

above 200:

taking 3 digit numbers the first number will be 2,7,8 3 cases are there and remaining 2 places are permutated at 4P2 = 12 ways

the first place will have 3 chances so... 3*12= 36

and the total numbers above 200 are 36+24=60 i hope...

Answer 3

The first part is right, but obviously the number of numbers over 200 can't be 60, since there are 16 numbers with fewer than 3 digits, and the total number of numbers is 64.
In actuality all of the 4-digit numbers will of course be bigger than 200. For the 3-digit numbers all except the numbers starting with 1 will be bigger than 200, which are 3/4 of the total since there are 4 numbers to choose from for the first digit. So the number of numbers over 200 is 24+24*3/4=42. Which gives us a probability of 42/64=21/32.

Answer 4

Thanks a lot! :-D