Given - The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.

To find - If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects

Proof -

Given that,

P(birth that result in a birth defect) = 1/33

P(birth that not result in a birth defect) = 1 - 1/33 = 32/33

Now,

Given that, n = 500

X = Number of birth that does not result in birth defects

Now,

P(X ≥ 490) = $\sum\limits^{500}_{x=490} {^{500} C_{x} } (\frac{32}{33} )^{x} (\frac{1}{33} )^{500-x}$

= ${^{500} C_{490} } (\frac{32}{33} )^{490} (\frac{1}{33} )^{500-490}$ + .......+ ${^{500} C_{500} } (\frac{32}{33} )^{500} (\frac{1}{33} )^{500-500}$

= 0.04541 + ......+0.0000002079

= 0.1076

⇒Probability that at least 490 do not result in birth defects = 0.1076

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2 years ago

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