SOLUTION: A fair decimal die is rolled three times. What is the probability that
All three outcomes represent odd numbers?
Answer: 1/8
The digit 1 occurs at least once?
Answer: 22/
Algebra.Com
Question 1193822: A fair decimal die is rolled three times. What is the probability that
All three outcomes represent odd numbers?
Answer: 1/8
The digit 1 occurs at least once?
Answer: 22/125
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!
Problem 1
I'm assuming a decimal die consists of the ten numbers 1,2,3,4,5,6,7,8,9,10
Though if instead it's 0,1,2,3,4,5,6,7,8,9, then the steps and answers mentioned below would be practically identical.
Half of them are odd (1,3,5,7,9) which means 1/2 is the probability of getting an odd result when rolling the die.
The probability of getting three odd numbers in a row is (1/2)*(1/2)*(1/2) = (1/2)^3 = 1/8
Each roll is independent of one another.
Answer: 1/8
You have the correct answer. Nice work.
=======================================================
Problem 2
Here's the sample space
{1,2,3,4,5,6,7,8,9,10}
which is the set of all possible outcomes.
Ask yourself: what is the probability of rolling anything but "1"?
The answer would be 9/10 since we have nine things we want (2 through 10) out of 10 outcomes total.
Stringing three such results together in a row would have the odds of (9/10)^3 = 729/1000
This represents the probability of rolling three items in a row, none of which are "1". Repeats are allowed.
The complement of this probability is
1 - (729/1000) = 271/1000
which represents the probability of the result "1" showing up at least once. This is because we have two options of either: 1 shows up at least once, or 1 doesn't show up at all.
Answer: 271/1000
I'm not sure how you got 22/125, so please let me know your steps and/or thought process. If your teacher wrote 22/125 as the answer, then s/he made a typo. If I made a mistake somewhere, then please let me know.
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