Question
Help me with this question please
There are 12 red checkers and 3 black checkers in a bag. Checkers are selected one at a time, with replacement. Each time, the color of the checker is recorded. Find the probability of selecting a red checker exactly 7 times in 10 selections. Show your work
There are 12 red checkers and 3 black checkers in a bag. Checkers are selected one at a time, with replacement. Each time, the color of the checker is recorded. Find the probability of selecting a red checker exactly 7 times in 10 selections. Show your work
Answers
binomial distribution
p(sucess) = 12/15 = 4/5 = .8
1-p = 1/5 = .2
n = 10
k = 7
P(x=7) = C(10,7) * .8^7 * .2^3
C(10,7) = 10!/[ 7! (3!) ]
= 10*9*8 / (3*2) = 80*9/6 = 120
so
P = (120)(.8^7)(.2^3) = 0.201
p(sucess) = 12/15 = 4/5 = .8
1-p = 1/5 = .2
n = 10
k = 7
P(x=7) = C(10,7) * .8^7 * .2^3
C(10,7) = 10!/[ 7! (3!) ]
= 10*9*8 / (3*2) = 80*9/6 = 120
so
P = (120)(.8^7)(.2^3) = 0.201
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(12/15)^7 * (3/15)^3 = ?
(12/15)^7 * (3/15)^3 = ?
....but the probability of failure is not independent of the probability of success :)
In other words you did one of 120 sequences that yield 7 successes out of 10 trials.
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