dice question

[GabeFried]

A pair of dice is rolled 12 times. The probability that doubles are rolled exactly 3 times is: a) 0.19740 b) 0.03842 c) 0.09529 d) none of the above

ive tried watching videos on this topic but im stuck and unsure how to start to solve this, any help is apreciated

 

[Deleted member 4993]

A pair of dice is rolled 12 times. The probability that doubles are rolled exactly 3 times is: a) 0.19740 b) 0.03842 c) 0.09529 d) none of the above

ive tried watching videos on this topic but im stuck and unsure how to start to solve this, any help is apreciated

How many outcomes are there for a pair of die being rolled once?I

How many ways a double can show up (2 did 1 roll)?

 

[GabeFried]

How many outcomes are there for a pair of die being rolled once?I

How many ways a double can show up (2 did 1 roll)?

so for 1 dice the P is 1/6 and for 2 it would be 1/36?

 

[pka]

A pair of dice is rolled 12 times. The probability that doubles are rolled exactly 3 times is: a) 0.19740 b) 0.03842 c) 0.09529 d) none of the above

First, this binominal probability : three successes and nine complements. There are six possible doubles on each roll of the dice.
Hence the probability of a double on any roll is \(\dfrac{6}{36}=\dfrac{1}{6}\)
Now apply binomial probability; \(\dbinom{12}{3}\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)^9\)

 

[GabeFried]

First, this binominal probability : three successes and nine complements. There are six possible doubles on each roll of the dice.
Hence the probability of a double on any roll is \(\dfrac{6}{36}=\dfrac{1}{6}\)
Now apply binomial probability; \(\dbinom{12}{3}\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)^9\)

i understand where you get 12c3 and 1/6 but am unsure where the ^3 and 5/6^9 comes from, my guess is that the 5/6 is the failures of it not being a double ?

 

[pka]

i understand where you get 12c3 and 1/6 but am unsure where the ^3 and 5/6^9 comes from, my guess is that the 5/6 is the failures of it not being a double ?

To GabeFried, you clearly did not read and understand the question: A pair of dice is rolled 12 times. (What is) probability that doubles are rolled exactly 3 times. Therefore three successes and nine complements.
In \(K\) trails the the probability of exactly \(0\le J\le K\) successes is \(\dbinom{K}{J}(p)^J(1-p)^{K-J}\)

 

[JeffM]

i understand where you get 12c3 and 1/6 but am unsure where the ^3 and 5/6^9 comes from, my guess is that the 5/6 is the failures of it not being a double ?

With n independent trials and a probability of success on a single trial of p, the probability of EXACTLY m successes is

[MATH]\dbinom{n}{m} * p^m * (1 - p)^{(n-m)}.[/MATH]
(1 - p) is the probability of a failure on a single trial because the probability of success or failure on a single trial = 1 and the probability of success and failure on a single trial = 0. And if we have m successes in n trials, we must have (n-m) failures as well.