A game consists of rolling a die 8 times. You win if the result is 5 or 6. Your numbers of wins will be recorded.
a) Identify the discrete random variable, X ans: Each time you role a die, if the result is 5 or 6, you win the game
b) State the number of trials, n
ans: As it is stated 8 is number of trials
c) State the probability of success, p, in any trial
ans: 1.52* 10^-4
d) State the probability of failure, q, in any trial.
ans: 0.9999
Hi I was doing prctice questions and saw this question, I answered them but not really sure if the answers are correct. I checked online , it did not help me much. I wouold appreciate it if you could help me see my answers are correct or not.
Having rolled the die eight times, we are interested in counting how many (if any) of those times a 5 or 6 was rolled.
We give a name to this "counting of how many 5's or 6's"... we call it a random variable and it is customary to name a random variable $X$.
That is to say, $X$ is a way of looking at the results of our eight rolls of dice and being able to say aloud "there were three 5's or 6's" or "there were no 5's and no 6's" etc...
$X$ has "support" $\{0,1,2,3,4,5,6,7,8\}$ as these are the possible answers to the question of "how many 5's or 6's did we roll in our 8 rolls of the dice."
It can be said that our random process consists of several smaller identical independent "trials"... each of which correspond to a single roll of a die. A "success" in one of these smaller trials correspond to us getting a 5 or 6 on the die.
The probability of success in a specific trial is going to simply be $\dfrac{1}{3}$ as it is assumed that the dice used are standard fair six-sided dice and two of the six sides correspond to a "win" here.
The probability of failure in a specific trial is going to simply be $\dfrac{2}{3}$, the opposite of the probability of success.
The answers as I would have written it: a) $X$ is the random variable counting the total number of successes across all trials. b) There are $n=8$ trials. c) The probability of success for a specific trial is $p=\dfrac{1}{3}$. d) The probability of failure for a specific trial is $q=1-p=\dfrac{2}{3}$
We call $X$ "binomially distributed" in this case and abbreviate all of the above into one statement, $X\sim\text{Bin}(8,\dfrac{1}{3})$
We can now talk about the probability that $X$ gives a certain value, i.e. that we rolled a certain number of 5's or 6's. We have that the probability of us rolling $k$ 5's or 6's will be:
$$\Pr(X=k)=\binom{8}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{8-k}$$
This information was not needed however for the parts of the question which you asked though it may likely appear in a followup question or parts of the question which you did not yet write.