sestigtalill

Answered

2022-12-19

Geometric distribution. Johnny has 1,176 Pok´emon cards in total. Pok´emon EX is a special type of card, and Johhny has 39 EX-type cards. He is looking for an EX-type card, but all of the cards are completely mixed up and stored in a shoe box. His mother is calling him for dinner. What is the probability that Johnny will have to look through no more than 25 cards before he finds an EX-type card?

Answer & Explanation

Agour2q9

Expert

2022-12-20Added 1 answers

$\text{Distribution: For}x=x\text{these must be a run of}(x-1)\text{failure followed by a success, so}\phantom{\rule{0ex}{0ex}}P(x=x)=\theta (1-\theta {)}^{x-1}\phantom{\rule{0ex}{0ex}}x=1,2,3,...(0\theta 1)\phantom{\rule{0ex}{0ex}}\text{I knows way of the geometric distribution let x be the numbers of failure's between the first success}\phantom{\rule{0ex}{0ex}}\text{than}P(x=x)=\theta (1-\theta {)}^{x}\phantom{\rule{0ex}{0ex}}x=0,1,2,3,...\phantom{\rule{0ex}{0ex}}P(x\le 25)\phantom{\rule{0ex}{0ex}}=P(x=0)+P(x=1)+--+P(x=24)\phantom{\rule{0ex}{0ex}}+P(x=25)\phantom{\rule{0ex}{0ex}}=\theta (1-\theta {)}^{0}+\theta (1-\theta {)}^{1}+(1-\theta {)}^{2}\phantom{\rule{0ex}{0ex}}+--+\theta (1-\theta {)}^{25}\phantom{\rule{0ex}{0ex}}=\theta [1+(1-\theta )+(1\theta {)}^{2}+--+(1-\theta {)}^{25}]\phantom{\rule{0ex}{0ex}}=\theta \ast \frac{[1-(1-\theta {)}^{26}]}{[1-(1-\theta )]}\phantom{\rule{0ex}{0ex}}=\frac{\theta}{\theta}[1-(1-\theta {)}^{26}]=1-(1-\theta {)}^{26}\phantom{\rule{0ex}{0ex}}\text{where}\theta =\frac{39}{1176}\phantom{\rule{0ex}{0ex}}P(x\le 25)=1-[1-\frac{39}{1176}{]}^{26}\phantom{\rule{0ex}{0ex}}1-{\frac{39}{1176}}^{26}\phantom{\rule{0ex}{0ex}}=0.5839$

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