Let N be a randomly chosen 5-digit number.
a. What is the probability that N contains at least one zero?
b. What is the probability that N contains at least one 7?
c.
What is the probability that it contains exactly one zero and one 7?
d. Generalize the above questions for a n-digit number , n>1.
See MathMan's correction to parts c and d.c. My solution here neglects the limitation on the possibilities for where the 7 goes imposed by the lack of zeros in the remaining four positions.
a. The first digit can't be zero. The probability that the rest of the digits are all non-zero is (9/10)^4. The probability that there's at least one zero is 1 - (9/10)^4 = .3439.
b. The probability that the first digit is not a 7 os 8/9. The probability that the rest of the digits are all non-7 is (9/10)^4. The probability that all are non-7 is 8/9 * (9/10)^4. The probability that there's at least one 7 is 1 - (8/9) * (9/10)^4 = .4168
c. The first digit is not under consideration for zero; it's not a zero. The probability that the second digit is a zero but not the others is (1/10)*(9/10)^3 = 729/10000 = .0729. This is multiplied by 4 for the four positions that the zero could be in, making .2916. Now that it has one zero any of the four remaining places could have the 7, so this is like trying to find exactly one zero among four positions--this time being one 7 among four positions. So the overall probability is .2916 ^ 2 = .08503056.
d.a. For an n-digit number the probability is 1 - (9/10)^(n-1).
d.b. 1 - 8/9 * (9/10)^(n-1)
d.c. For an n-digit number this is (((n-1)/10)*(9/10)^(n-2))^2
Edited on July 8, 2018, 8:27 pm
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Posted by Charlie
on 2018-06-28 12:26:17 |
solution
