One thousand independent rolls of a fair die is made.

conditional probabilityprobabilityprobability distributionsprobability theory

Question: One thousand independent rolls of a fair die is made. If the number $6$ appears exactly $200$ times, find the probability that number $5$ appears less than $150$ times. (Use appropriate statistical table)

Answer: $0.1762$

Let $p$ be the probability of rolling a 6 on one roll, and $q=1-p$.

Now this is a problem of conditional probability where we have to apply binomial approximation to standard normal distribution since $n \to \infty$ and $p,q < \infty$.

Let $X_1$ be the random variable corresponding to the getting of $6$ and $X_2$ be the random variable corresponding to the getting of $5$.

Now required probability is
$$
P(X_2<150|X_1=200)
=\frac{P((X_2<150 )\ \mathrm{ and }\ (X_1=200))}{P(X_1=200)}
$$

Now the denominator can be easily found but I face problem with the numerator.

How to proceed ?

Best Answer

In this case is quite trivial to compute the conditional probability: you made 1000 rolls and you know 200 of those were not a 5. Also, the remaining rolls were picked from the set {1,2,3,4,5}. Hence, it is the same as rolling a 5-sided die 800 times.

$P(X_2 < 150)$ is then approximated by a normal distribution with mean $\mu = 800/5 = 160$ and $\sigma^2 = 800(1/5)(4/5) = 128$. EDIT: As it was pointed out, you should take 0.5 from 150, to account for the continuum approximation from a discrete space.

If you are to use a table, you should now standarize your distribution, which is done by taking the mean from $X_2$ and then dividing by $\sigma$. Hence, what you want to compute is

$$P(X_2 < 150) = \int_{-\infty}^{\frac{149.5-160}{\sqrt(128)}} \frac{1}{2\pi} exp(-\frac{1}{2}x^2)= 0.1762 $$

according to WolframAlpha.

Again, if you are looking on a table, usually they give integrals from $-\infty$ to some positive number. Since you look for the integral from $-\infty$ to -0.928 aprox, which is negative, this is the same as integrating from +0.928 to $+\infty$, or otherwise, 1 minus the probability integrated from $-\infty$ to +0.928.

So look on the table for the closest thing to 0.928, name that $p$ and then the answer is $1-p$.

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