SOLUTION: The probability of passing a class is 60%. What is the probability that 1 out of 20 students pass the class? A) 1.2*10^-4% B) 2.3*10^-4% C) 3.3*10^-5% D) 5.6*10^-5%

SOLUTION: The probability of passing a class is 60%. What is the probability that 1 out of 20 students pass the class? A) 1.210^-4% B) 2.310^-4% C) 3.310^-5% D) 5.610^-5%

Question 1082468: The probability of passing a class is 60%. What is the probability that 1 out of 20 students pass the class?
A) 1.2*10^-4%
B) 2.3*10^-4%
C) 3.3*10^-5%
D) 5.6*10^-5%
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the formula for the binomial probability is:

p(x) = p^x * q^(n-x) * c(n,x)

when p = .6, q = .4 because q = 1 - p.

when x = 1 and n = 20, the formula becomes:

p(1) = .6^1 * .4^19 * c(20,1)

c(20,10) is the combination formula of c(n,x) = n! / (x!*(n-x)!)

when n = 20 and x = 1, this formula becomes c(20,1) = 20! / (1! * 19!)

this can be written as c(20,1) = (20 * 19!) / (1! * 19!)

simplify this to get c(20,1) = 20/1 = 20

p(1) = .6^1 * .4^19 * c(20,1) becomes p(1) = .6^1 * .4^19 * 20.

solve for p(1) to get p(1) = 3.298534883 * 10^(-7)

round this to 1 decimal digit and this becomes p(1) = 3.3 * 10^(-7)

multiply this by 100 to make it a percent and you get p(1) = 330 * 10^(-7)%

convert this to scientific format to get p(1) = 3.3 * 10^(-5)%

to convert to scientific format, there must be only 1 digit to the left of the decimal point

330 * 10^(-7)% has 3 digits to the left of the decimal point because the decimal point is implied after the 330.

it actually looks like 330. * 10^(-7)%

to get it to be 3.3, you have to divide 330 by 100 and multiply 10^(-7) by 100.

you would get p(1) = 3.3 * 10^(-7) * 100

10^(-7) * 100 is the same as 10^(-7) * 10^2 which results in 10^(-7+2) which results in 10^(-5)

you end up with p(1) = 3.3 * 10^(-5)%.

that looks like selection C.

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