A wheel of fortune is divided into 10 parts, one of them brings the jackpot.

[mathmari]

Hey!

A wheel of fortune is divided into $10$ equal sized parts, of which one of them brings the jackpot.

• A player wants to investigate the regularity of the wheel of fortune. He turns the wheel 20 times. Calculate the probability that he will get at least one main prize if it is a laplace wheel.
• How often does the player have to spin the wheel of fortune to get at least one main prize with a probability of at least 90%.
• The player objects to the wheel when less than twice the jackpot appears at 20 times turn. Calculate the probability that the wheel will be misjudged.
• What does the player have to change in his test so that the significant level is about $6\%$ ?

I have done the following:

• $$P(X\geq 1)=1-P(X\leq 1)=1-\sum_{i=0}^1\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}$$
  
  $$ $$
  
• \begin{align*}P(X\geq k)\geq 90\% &\Rightarrow 1-P(X\leq k)=90\% \\ & \Rightarrow 1-\sum_{i=0}^k\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}\geq 0.9 \\ & \Rightarrow \sum_{i=0}^k\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}\leq 0.1\end{align*}

Is everything correct so far? (Wondering)

Could you give me a hint for the other two questions? (Wondering)

 

[I like Serena]

Hey!

A wheel of fortune is divided into $10$ equal sized parts, of which one of them brings the jackpot.

• A player wants to investigate the regularity of the wheel of fortune. He turns the wheel 20 times. Calculate the probability that he will get at least one main prize if it is a laplace wheel.
• How often does the player have to spin the wheel of fortune to get at least one main prize with a probability of at least 90%.
• The player objects to the wheel when less than twice the jackpot appears at 20 times turn. Calculate the probability that the wheel will be misjudged.
• What does the player have to change in his test so that the significant level is about $6\%$ ?

I have done the following:

• $$P(X\geq 1)=1-P(X\leq 1)=1-\sum_{i=0}^1\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}$$

Hey mathmari! (Smile)

Shouldn't it be:
$$P(X\geq 1)=1-P(\text{NOT }(X\geq 1))=1-P(X<1)$$
(Wondering)

• \begin{align*}P(X\geq k)\geq 90\% &\Rightarrow 1-P(X\leq k)=90\% \\ & \Rightarrow 1-\sum_{i=0}^k\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}\geq 0.9 \\ & \Rightarrow \sum_{i=0}^k\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}\leq 0.1\end{align*}

Is everything correct so far?

You are one off. (Sweating)

Next thing to do is calculate those chances for $i=0$, $i=1$, and so on until we reach a sum of $0.1$. (Thinking)

Or alternatively we can use a tool or table to find the inverse cumulative binomial distribution. (Nerd)

Could you give me a hint for the other two questions?

Suppose the wheel really is a laplace wheel.
We will call this the null hypothesis ($H_0$).
Then we can calculate the probability that the player will object anyway, can't we?
That is, we can calculate $P(X \le 2\mid H_0)$, can't we? (Wondering)This probability that we misjudge even though the null hypothesis is true, is called the significance or $p$-value.
If just now we found that the probability to misjudge is more than $6\%$, what can we do to change that probability? (Wondering)