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Algebra 19 Online
OpenStudy (anonymous):

The probability of a batch of 200 vaccines containing 0, 1, 2, 3, 4, or 5 ineffective vaccines is given in the following probability distribution. #16Q1.png What is the probability that a random batch of 200 vaccines contains at least 2 ineffective vaccines? a. 0.20 b. 0.35 c. 0.65 d. 0.80

geerky42 (geerky42):

Try to attach this picture #16Q1.png

OpenStudy (anonymous):

OpenStudy (kirbykirby):

At least 2 means \(P(X \ge 2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)\)

OpenStudy (kirbykirby):

omg please stop with the anne pic >_<

OpenStudy (kirbykirby):

ya I saw you deleted your reply

OpenStudy (anonymous):

ok...so what am I supposed to do?

OpenStudy (kirbykirby):

add up the probabilities ... i.e the y-value on your bar graph for x=2, x=3, x=4, x=5

OpenStudy (anonymous):

@kirbykirby I add 0.05+0.1+0.15 etc? and then?

OpenStudy (kirbykirby):

0.15, 0.10, 0.05 and 0.05

OpenStudy (anonymous):

I am confused -.-"

OpenStudy (kirbykirby):

Add the probabiltiies for ineffectiveness at x=2, x=3, x=4, x=5

OpenStudy (kirbykirby):

At x=2, the probability s 0.15 at x=3, the probability is 0.10 at x =4, the probability is 0.05

OpenStudy (kirbykirby):

at x=5 the probability is 0.05

OpenStudy (anonymous):

got that.

OpenStudy (anonymous):

I guess it's 0.35?

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