Probability

Question

Student is assigned 4 digit pin to access money in personal lunch account.  Code is randomly generated and first digit cannot be 0. What is the probability that a student lunch code will be a perfect square?

Answer 1

the total  number of possible 4-digit pins is 9*10*10*10=9000

for the first digit you are picking from the digits 1-9 (first digit can't be 0)

for the second, third, and fourth digits you are picking from the digits 0-9

 

the perfect square has to be a 4-digit number

√1000=31.6 so the first perfect square has to be 32²=1024

the last perfect square has to be less than 10,000 because 10,000 is a 5-digit number

the last perfect square has to be less than 100² which equals 10,000

the last perfect square must be 99²=9801

the perfect squares are 32², 33², 34²,... 97², 98², and 99²

from 32 to 99 there are 68 numbers because 1 to 99 gives us 99 numbers but we are leaving out the numbers 1-31

99-31=68

probability is total number of favorable outcomes divided by total number of possible outcomes

probability of a perfect square is 68/9000=34/4500=17/2250

 

Answer 2

how many numbers between 1000 and 9999 are perfect squares? call this x

How many 4 digit numbers are possible? call this y

divide x/y to get probability.

 

I'll let you find x. (HINT: the square root of 10,000 is 100 and the square root of 1024 is 32)

 

To get y

 

There are 9 possibilities for the first digit, and 10 for the second, third and fourth.

y = 9 x 10 x 10 x 10 = 9000