A spinner is spun 8 times. What is the probability of landing on a 3 more than 5 times, if the probability of success is 0.2?
I used this expression to solve for it, but I keep getting the wrong answer! (8C6 is 8 composition 6)
8C6 (0.2^6) (0.8^2) + 8C7 (0.2^7) (0.8^1) + 8C8 (0.2^8) (0.8^0) = 0.0012
But the correct answer is 0.010?
I used this expression to solve for it, but I keep getting the wrong answer! (8C6 is 8 composition 6)
8C6 (0.2^6) (0.8^2) + 8C7 (0.2^7) (0.8^1) + 8C8 (0.2^8) (0.8^0) = 0.0012
But the correct answer is 0.010?
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The problem follows a binomial distribution, the formula of which is:
∑(i=0,x) [(nCi)(p^i)(1 − p)^(n − i)]
As you can see, the summation starts with i = 0, not i = 1. That means the fifth term will be i = 4.
P(X > 5)
= 1 − P(X ≤ 5)
= 1 − [(8C0)(.2)^0(.8)^8 + (8C1)(.2)^1(.8)^7 + ... + (8C4)(.2)^4(.8)^4
= 1 − .9895936
= .0104064
∑(i=0,x) [(nCi)(p^i)(1 − p)^(n − i)]
As you can see, the summation starts with i = 0, not i = 1. That means the fifth term will be i = 4.
P(X > 5)
= 1 − P(X ≤ 5)
= 1 − [(8C0)(.2)^0(.8)^8 + (8C1)(.2)^1(.8)^7 + ... + (8C4)(.2)^4(.8)^4
= 1 − .9895936
= .0104064
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Hmm..I'm getting the same answer. Maybe your answer sheet is wrong?