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In each month, the proportion of ‘Prize’ bonds that win a prize is 1 in 11,000. There is a large number of prizes and all bonds are equally likely to win each prize.

(a) Show that, for a given month, the probability that a bondholder with
5000 bonds wins at least one prize, is 0.365, correct to 3 significant figures.

(b) For a given month, find the probability that, in a group of 10 bondholders
each holding 5000 bonds, four or more win at least one prize

Respuesta :

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Answer:

Step-by-step explanation:

Given that:

The proportion of price bonds that wins a prize is 1 in 111000

The probability of winning at least one prize = 1 - Probability of winning no prize at all.

Probability of winning at least one prize = [tex]1 - \bigg (\dfrac{10999}{11000} \bigg)^{5000}[/tex]

Probability of winning at least one prize = 1 - 0.9999090909⁵⁰⁰⁰

Probability of winning at least one prize = 1 - 0.635

Probability of winning at least one prize = 0.365

Since this question follows a binomial distribution with the probability of p = 0.365 and number of sample n = 10

Then; the probability of four or more will win at least one prize can be computed as follows:

[tex]= \sum \limits ^{10}_{x=4} \bigg( ^{10}_{x} \bigg)(0.365)^x (1- 0.365)^{10-x}[/tex]

[tex]= \bigg( ^{10}_{4} \bigg)(0.365)^4 (0.635)^{10-4}[/tex]

[tex]= \bigg( \dfrac{10!}{4!(10-4)!} \bigg)(0.365)^4 (0.635)^{6}[/tex]

[tex]= 0.244[/tex]

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