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Fiesta28 [93]
3 years ago
12

Lotteries are an important income source for various governments around the world. However, the availability of lotteries and ot

her forms of gambling have created a social problem: gambling addicts. A critic of government-controlled gambling contends that 30% tickets are gambling addicts. If we randomly select 10 people among those who report that they regularly buy lottery tickets, What is the probability that more than 5 of them are addicts?
Mathematics
1 answer:
SashulF [63]3 years ago
5 0

Answer: 0.0473

Step-by-step explanation:

Given : The proportion of gambling addicts : p=0.30

Let x be the binomial variable that represents the number of persons are gambling addicts.

with parameter p=0.30 , n= 10

Using Binomial formula ,

P(x)=^nC_xp^x(1-p)^{n-x}

The required probability = P(x>5)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\=^{10}C_6(0.3)^6(0.7)^{4}+^{10}C_7(0.3)^7(0.7)^{3}+^{10}C_8(0.3)^8(0.7)^{2}+^{10}C_9(0.3)^9(0.7)^{1}+^{10}C_{10}(0.3)^{10}(0.7)^{0}\\\\=\dfrac{10!}{4!6!}(0.3)^6(0.7)^{4}+\dfrac{10!}{3!7!}(0.3)^7(0.7)^{3}+\dfrac{10!}{2!8!}(0.3)^8(0.7)^{2}+\dfrac{10!}{9!1!}(0.3)^9(0.7)^{1}+(1)(0.3)^{10}\\\\=0.0473489874\approx0.0473

Hence, the required probability = 0.0473

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