A person buys a lottery ticket in 50 lotterie | vedclass

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Question

(A) Let $X$ represent the number of winning prizes in $50$ lotteries. The trials are Bernoulli trials. Clearly,$X$ has a binomial distribution with $n

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $X$ represent the number of winning prizes in $50$ lotteries. The trials are Bernoulli trials. Clearly,$X$ has a binomial distribution with $n = 50$ and $p = \frac{1}{100}$.$	herefore q = 1 - p = 1 - \frac{1}{100} = \frac{99}{100}$.$	herefore P(X = x) = ^{n}C_{x} q^{n-x} p^{x} = ^{50}C_{x} \left(\frac{99}{100}ight)^{50-x} \left(\frac{1}{100}ight)^{x}$.$P(	ext{at least twice}) = P(X \geq 2) = 1 - P(X $P(X = 0) = ^{50}C_{0} \left(\frac{99}{100}ight)^{50} \left(\frac{1}{100}ight)^{0} = \left(\frac{99}{100}ight)^{50}$.$P(X = 1) = ^{50}C_{1} \left(\frac{99}{100}ight)^{49} \left(\frac{1}{100}ight)^{1} = 50 \cdot \left(\frac{99}{100}ight)^{49} \cdot \frac{1}{100} = \frac{1}{2} \left(\frac{99}{100}ight)^{49}$.$P(X \geq 2) = 1 - \left[ \left(\frac{99}{100}ight)^{50} + \frac{1}{2} \left(\frac{99}{100}ight)^{49} ight]$.$= 1 - \left(\frac{99}{100}ight)^{49} \left[ \frac{99}{100} + \frac{1}{2} ight] = 1 - \left(\frac{99}{100}ight)^{49} \left[ \frac{99 + 50}{100} ight] = 1 - \left(\frac{149}{100}ight) \left(\frac{99}{100}ight)^{49}$.

$A$ person buys a lottery ticket in $50$ lotteries,in each of which his chance of winning a prize is $\frac{1}{100}$. What is the probability that he will win a prize at least twice?

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