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PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!-example-1

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The probability of 2 success among 5 trials is 0.132.

Solution:

Given n = 5, x = 2, p = 0.70

Binomial distribution formula:

The probability of x success in n trials is


P(X)={C^n_x}\ p^x\ q^(n-x)

where p is the probability of success and q is the probability of failure,

q = 1 –p

= 1 – 0.70

q = 0.30

Therefore, the probability of 2 success among 5 trials is


P(X)={C^5_2}\ (0.70)^2\ (0.30)^(5-2)


$=(5* 4)/(1 * 2) * \ (0.70)^2\ *(0.30)^3


$=10 * \ 0.49* 0.027


P(X)=0.1323

P(X) ≈ 0.132 (Round to three decimal)

Hence the probability of 2 success among 5 trials is 0.132.

User Metatoaster
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