# Probability that last two draws of integer product will be odd.

An urn contains 6 balls, each labeled differently with one of the numbers from 1 to 6. Three balls are extracted without replacement.

a. What is the probability that the absolute value of the difference between the first number extracted and the second is greater or equal than 4?

b. What is the probability that the product of the second number extracted and the third is an odd number?

• With such a small collection you can just do it case by case. Form part a: Suppose the first draw is $1$...then you need the second to be $5,6$. Now suppose the first draw is $2$...then you need the second to be $6$. And so on. – lulu Dec 11 '18 at 12:35
• I understand that there are 6 possible ways to have an absolute value greater or equal than 4: (6, 1), (1, 6), (6, 2), (2, 6), (5, 1), (1, 5). However, I'm not sure how to proceed from there ... – Ulrich Paul Wohak Dec 11 '18 at 13:16
• Ok, and how many possible ways can you draw the first two numbers? – lulu Dec 11 '18 at 13:19
• Correct. Well, a technical point: $\frac 6{30}=\frac 15$ is the probability. The "odds" are not written the same way. Here the odds would be $6:24$ as there are $6$ winning outcomes and $24$ losing outcomes. – lulu Dec 11 '18 at 13:27
• Seems overly complicated. There are $\binom 32=3$ ways to choose a pair of odd numbers from $\{1,\cdots, 6\}$. There are $\binom 62=15$ ways to choose a a pair of numbers from the list. Thus the probability is $\frac 3{15}=\frac 1{5}$. – lulu Dec 11 '18 at 13:39