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makvit [3.9K]
3 years ago
10

Suppose that on each play of a game, a gambler either wins 1 with probability p or loses 1 with probability 1–p (or q). The gamb

ler keeps betting until she or he is either up a total of n or down a total of m. What is the probability the gambler will quit an overall winner? You must consider both cases when p = 0.5 and when p ≠ 0.5
Mathematics
1 answer:
musickatia [10]3 years ago
3 0

Answer:

Step-by-step explanation:

From the given information,

Considering both cases when p = 0.5 and when p ≠ 0.5

the probability that the gambler will quit an overall winner is:

P = \dfrac{1 - (\dfrac{1-p}{p} )^K}{1- (\dfrac{1-p}{p})^N } \ \ \  is \  p  \neq 0.5  \  and\  K/N = \dfrac{1}{2}

where ;

N.k = n  and k  = m

Hence, the probability changes to:

P = \dfrac{1 -(\dfrac{1-p}{p})^m}{1 -(\dfrac{1-p}{p})^{m+n}} is  p ≠ 0.5   and k/N = \dfrac{m}{m+n}  is P = 0.5

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F<span>or every 1,000 feet you go up in elevation</span>, <span>the temperature decreases by about 3.3°F</span>

 

Given:

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Step-by-step explanation:

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Answer:

20

Step-by-step explanation:

Use <u>PEMDAS</u>

P = parenthesis

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M = Multiplication*

D = Division*

A = Addition**

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*either can come first, it just depends which comes first in the equation.

**either can come first, it just depends which comes first in the equation.

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Read 2 more answers
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