0
$\begingroup$

A die is continuously rolled 105-1 times. What is the probability that the total sum of all rolls does not exceed 400.?

I'm sorry if the question looks weird. I copied it the exact way I saw it. Is the 105-1 a convention that I'm not aware of?

I have tried solving the problem by subtracting the possibility of the total sum exceeding 400 from 1 but got the wrong answer.

  • 0
    If you are copying questions on to this website, then I suggest you add some context to the question, like your attempts on solving this question, your interpretation of the question, where you copied it from etc. This will help us give better answers to your question. Having said that, I don't know what $105-1$ means. – 2017-01-18
  • 0
    $105 - 1$ is simply $104$. Maybe there's some significance to the number $105$ in the source of the question. – 2017-01-18
  • 1
    Also the term "continuously" is strange here, and perhaps a mistake for "repeatedly". – 2017-01-18
  • 0
    Once you know how many times you are tossing the die, I'd suggest using the normal approximation with mean $N\times \frac 72$ and variance $N\times \frac {35}{12}$. – 2017-01-18

1 Answers 1

0

Let $ X_i $ denote the value of the die on the $ i^{th} $ roll. We want to find $P\big(\sum_{i=1}^{104} X_i\big) \leq 400 $.

As @lulu pointed out in the comments section, you want to use the CLT, i.e.

$$ \frac{\sum_{i=1}^{104}X_i - E(\sum_{i=1}^{104}X_i)}{\sqrt{Var(\sum_{i=1}^{104}X_i)}} \rightarrow Z $$

Since $E(X) = \frac{1}{6}\big(1 + 2 + 3 + 4 + 5 + 6\big) = \frac{7}{2}$ and $Var(X)=E(X^2)-(E(X))^2 = \frac{1}{6}\big(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2\big) - \Big(\frac{7}{2}\Big)^2 = \frac{35}{12} $, we get:

$$P\Big(\sum_{i=1}^{104} X_i \leq 400\Big) = P\Bigg(\frac{\sum_{i=1}^{104}X_i - E(\sum_{i=1}^{104}X_i)}{\sqrt{Var(\sum_{i=1}^{104}X_i)}} \leq \frac{400 - E(\sum_{i=1}^{104}X_i)}{\sqrt{Var(\sum_{i=1}^{104}X_i)}}\Bigg) \approx P\Bigg(Z \leq \frac{400 - 104E(X)}{\sqrt{104Var(X)}}\Bigg) \approx P\Bigg(Z \leq \frac{400 - 104E(X)}{\sqrt{104Var(X)}}\Bigg) \approx P\Bigg(Z \leq \frac{400 - 104(7/2)}{\sqrt{104(35/12)}}\Bigg) \approx P(Z \leq 2.07) \approx 0.05 $$

Note: To improve the precision of your answer, you could also apply the continuity correction, which would mean calculating $P(\sum_{i=1}^{104} \leq 400.5)$ instead of $P(\sum_{i=1}^{104} \leq 400)$.