Thread: Restricted outcomes for die thrown six times.

Hi guys, need help with a question:

An unbiased die is thrown 6 times. Calculate the probabilities that the six scores obtained will be such that a 6 occurs only on the last throw and that exactly three of the first five throws result in odd numbers.

I assume the bits in blue are seperate questions, ie you are not expected to find the chance that exactly 3 of the first 5 are odd and the only 6 occurs at the last throw.

6 occurs only on the last throw

N = not a 6
S = 6

you want NNNNNS

exactly three of the first five throws result in odd numbers.

you can use the binomial distribution.


Post your attempt if more info is needed.

Hello, Blizzardy!

An unbiased die is thrown 6 times.
Calculate the probability that the six scores obtained will be such that:
. . a 6 occurs only on the last throw and that
. . exactly three of the first five throws result in odd numbers.

In the first 5 throws, we want 3 odd numbers and 2 even numbers.
. . The probability is: .

On the sixth throw, we want a "6".
. . The probability is


Answer: .

Originally Posted by Blizzardy

Hi guys, need help with a question:

An unbiased die is thrown 6 times. Calculate the probabilities that the six scores obtained will be such that a 6 occurs only on the last throw and that exactly three of the first five throws result in odd numbers.

Hi Blizzardy,

I think Soroban made a little slip in that he did not take into account the restriction that the *first* 6 must occur on the last throw. So here is an attempt to take that into account.

First, let's find the probability that the first five throws contain exactly three even numbers and two even numbers, none of which are a six. There are possible sequences of five numbers 1-6, all of which we assume are equally likely. How many of these have three odd numbers and two even numbers, excluding six? There are ways to select the positions for the odd and even numbers. Once we have chosen the positions, there are ways to pick the odd numbers and ways to pick the even numbers since we are excluding the six. So the total number of acceptable arrangements for the first five throws is

and the probability of getting five throws with three odd numbers and two even numbers other than six is


Then the probability that the final throw is a six is 1/6, so the answer to the problem is

Originally Posted by awkward

I think Soroban made a little slip in that he did not take into account
the restriction that the *first* 6 must occur on the last throw.

Absolutely right!

Thanks for catching that . . .