The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7?
Kaplan’s solution says:
If the range is 7, then we want 2 of the numbers to differ by 7 and the other 2 numbers to both be between the 2 numbers that differ by 7. E.g. if the numbers are 1 and 8, then the other two numbers could be only 2,3,4,5,6 or 7.
From what I’m reading, the problem does not state explicitly that the other 2 numbers must lie inside the range. Am I right or is there something I’m missing?
If one of the other two numbers is outside the range, the range will be bigger - you have to have the smallest and largest number, with the other two being in between. In the example, if the numbers are 1 and 8 and one of the other two numbers is 9, the range is actually 8, not 7.
I have another question about probabilities, the problem is:
Ale has a 6-sided die with faces from 1 to 6 and he rolls the dice twice. What is the probability that neither roll is a multiple of 3?
Approach 1:
From options 1,2,3,4,5,6 total 6
possible numbers not multiples of 3: 1,2,4,5 total 4 => (4/6).(4/6) = 4/9
Approach2:
To calculate the (1-prob) option, first both times roll a multiple of 3: 3 or 6
then: 3,6 (2) of 6 possibilities => (2/6).(2/6) = 1/9
Then rest: 1- 1/9 = 8/9 to have the non multiples of 6.
There must be a missing idea in my calculations, because the results should be the same in either approach. Could you please help me out to find the error here?
Your first approach is correct. In your second approach you first mistakenly calculated the probability of rolling a multiple of three in both rolls, not in either roll.
Thanks! I found the mistake, then the 2nd approach should be:
(1/9) + (2/6).(4/6).2 = 5/9 => 1 - 5/9 = 4/9
I guess as a non-native english speaker I get grammatically confused sometimes. In fact, there are some problems that use the words “at least”, then doing the opposite of what is asked (or taking the 1-prob approach) is more straight forward. Is there any clue to recognize with approach is more effective to time limitations?
Hmm, I don’t really have much advice - it was so long ago since I took probability and statistics. I usually try to follow your first approach, since I hate adding fractions!
