Balls

We have n identical balls (numbered 1-n) selected without replacement. Determine
1) The probability that at least one tensile strength number coincides with the number of balls?
2) Determine the mean and variance of the number of balls, which coincides with the number of balls in numbered order.

Final Answer:

p =  0.6333
m =  1
σ =  1

Step-by-step explanation:

$$\begin{gathered} p_5=5!=120 \\ {d}_5=p_5\cdot (1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})=120\cdot (1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})=44 \\ p=1-\frac{d_5}{p_5}=1-\frac{44}{120}=\frac{1\cdot 120}{120}-\frac{44}{120}=\frac{120}{120}-\frac{44}{120}=\frac{120-44}{120}=\frac{76}{120}=0.6333 \end{gathered}$$

$m=1$

$\sigma =1$

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