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As reported in Trends in Television, the proportion of US households who have at least one VCR is 0.592. If 8 households are selected at random, without replacement, from all US households, what is the (approximate) probability that the number having at least one VCR is exactly 4. Be sure to use many decimal places in your calculations (at least 4), but report your answer to three decimal places.

Sagot :

Fichohidn

Answer:

0.23825

Step-by-step explanation:

Given :

p = 0.592 ; number of trials, n = 8

x = 4 ; q = 1 - p = 1 - 0.592 = 0.408

Using the binomial probability relation :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 4) = 8C4 * 0.592^4 * 0.408^4

P(x = 4) = 70 * 0.592^4 * 0.408^4

P(x = 4) = 0.23825

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