Kelly Nelson
2021-12-19
Answered

If you buy a lottery ticket in 50 lotteries, in each of which your chance of winning a prize is $\frac{1}{100}$ , what is the (approximate) probability that you will win a prize (a) at least once, (b) exactly once, and (c) at least twice?

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alexandrebaud43

Answered 2021-12-20
Author has **36** answers

Step 1

$P\{X=i\}\approx {e}^{-\lambda}\frac{{\lambda}^{i}}{i!}$

Given:$n=50$

$p=\frac{1}{100}=0.01$

So,$\lambda =np=50\times 0.01=0.5$

a. Probability that you will win a prize at least once

$P\{X\ge 1\}\approx 1-P\{X=0\}=1-{e}^{-0.5}\frac{{0.5}^{0}}{0!}=0.3935$

b. Probability that you will win a prize exactly once

$P\{X=0\}\approx {e}^{-0.5}\frac{{0.5}^{1}}{1!}=0.3033$

c. Probability that you will win a prize at least twice

$P\{X\ge 2\}\approx 1-P\{X=0\}-P\{X=1\}=1-{e}^{-0.5}\frac{{0.5}^{0}}{0!}-{e}^{-0.5}\frac{{0.5}^{1}}{1!}=0.0902$

Given:

So,

a. Probability that you will win a prize at least once

b. Probability that you will win a prize exactly once

c. Probability that you will win a prize at least twice

Jonathan Burroughs

Answered 2021-12-21
Author has **37** answers

Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.

Clearly, X has a binomial distribution with$n=50\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}p=\frac{1}{100}$

$\therefore q=1-p=1-\frac{1}{100}=\frac{99}{100}$

$\therefore (X=x){=}^{n}{C}_{x}{q}^{n-x}{p}^{x}{=}^{50}{C}_{x}{\left(\frac{99}{100}\right)}^{50-x}\cdot {\left(\frac{1}{100}\right)}^{x}$

a) P (winning at least once)$=P(X\ge 1)$

$=1-P\left(X<1\right)$

$=1-P(X=0)$

$=1{-}^{50}{C}_{x}{\left(\frac{99}{100}\right)}^{50}$

$=1-1\cdot {\left(\frac{99}{100}\right)}^{50}$

$=1-{\left(\frac{99}{100}\right)}^{50}$

b) P (winning exactly once)$=P(X=1)$

$=}^{50}{C}_{1}{\left(\frac{99}{100}\right)}^{49}\cdot {\left(\frac{1}{100}\right)}^{1$

$=50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}$

$=\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}$

c) P (at least twice)$=P(X\ge 2)$

$=1-P\left(X<2\right)$

$=1-P(X\le 1)$

$=1-[P(X=0)+P(X=1)]$

$=[1-P(X=0)]-P(X=1)$

$=1-{\left(\frac{99}{100}\right)}^{50}-\frac{1}{2}\cdot {\left(\frac{99}{100}\right)}^{49}$

$=1-{\left(\frac{99}{100}\right)}^{49}[\frac{99}{100}+\frac{1}{2}]$

$=1-{\left(\frac{99}{100}\right)}^{49}\cdot \left(\frac{149}{100}\right)$

Clearly, X has a binomial distribution with

a) P (winning at least once)

b) P (winning exactly once)

c) P (at least twice)

Answered 2021-12-27

Let X:Number of times he wins a prize

Winning a prize on lottery is a Bernoulli trial

So, X has a binomial distribution

Here,

a) Probability that he wins the lottery atleast once

b) Probability that he wins the lottery exactly once

c)

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then

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I'm not understanding how theses equate to each other, why is it greater than equal to?

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I'm not understanding how theses equate to each other, why is it greater than equal to?

$$Pr(Y\le k)-P(Y\le l-1)=Pr({N}_{k}\ge r)-Pr({N}_{l-1}\ge r)$$

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Click here to view the standard normal distribution table (page 1).

LOADING...

Click here to view the standard normal distribution table (page 2).

LOADING...

(a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 160 students, at least 79 would prefer a boy, assuming the true percentage is 45%.

The probability that at least 79 students would prefer a boy is 3636.

(Round to four decimal places as needed.)

Click here to view the standard normal distribution table (page 1).

LOADING...

Click here to view the standard normal distribution table (page 2).

LOADING...

(a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 160 students, at least 79 would prefer a boy, assuming the true percentage is 45%.

The probability that at least 79 students would prefer a boy is 3636.

(Round to four decimal places as needed.)

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