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3 years ago

7

The probability of winning a particular game at the Game Place is 14 1/4 . Find the odds against winning the game.

1 answer:

Zinaida [17]3 years ago

6 0

ANSWER

$\frac{1}{3}$

EXPLANATION

We want to find an odds ratio from a given probability, which is

$\frac{1}{4}$

We subtract the numerator 1 from the denominator 4 to obtain

$4- 1= 3$

The answer is the number of unfavorable outcomes.

The Odds can then be expressed as

$1 : 3$

The ratio of favourable outcomes to unfavorable outcomes.

Or

$\frac{1}{3}$

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Answer:

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Step-by-step explanation:

Given that

$f'(x) = -\frac{1}{1 + x^{2}}$ ---(1)

We know that:

$\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}}$ ---(2)

Comparing (1) and (2)

$f'(x)=-(tan^{-1}x)$ ---- (3)

Using power series expansion for $tan^{-1}x$

$f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx$

$= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]$

$=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}$

$=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

as

$tan^{-1}(0)=0 \implies C=0$

Hence,

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

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