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A certain medical test is known to detect 59% of the people who are afflicted with the disease Y. If 10 people with the disease are administered the test, what is the probability that the test will show that:

Sagot :

MrRoyalidn

Answer:

[tex]P(x = 3) = 0.048[/tex]

Step-by-step explanation:

Given

[tex]n = 10[/tex]

[tex]p=59\% = 0.59[/tex]

Required

[tex]P(x = 3)[/tex] --- probability that 3 are afflicted

This question illustrates binomial probability and it is calcuated using:

[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]

So, we have:

[tex]P(x = 3) = ^{10}C_3 * 0.59^3 * (1 - 0.59)^{10-3}[/tex]

[tex]P(x = 3) = ^{10}C_3 * 0.59^3 * 0.41^7[/tex]

[tex]P(x = 3) = 120 * 0.59^3 * 0.41^7[/tex]

[tex]P(x = 3) = 0.048[/tex]

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