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What is the probability that a machine with 7 components that function independently with a failure rate of 2 what is the probability that at least 4 failed?
probability of a machine component failing = 2/7 P(at least 4 failed) = P( 4 failed)+P(5 failed) +P(6 failed)+P(7 failed) Using binomial probability: P(4 failed ) =7C4 (2/7)^4 ((5/7)^3 = 0.084987 P(5 failed) = 7C5 (2/7)^5 (5/7)^2 = 0.020395 P(6 failed ) = 7C6 (2/7)^6 (5/7) = 0.002719 P(7 failed) = (2/7)^7 = 0.000155 Adding, P(at least 4 failures) = 0.108257
If you have 6 sided die and a coin What is the probability that you roll a 5 and the coin lands on heads?
First, note that one even is independent of the other. If A and B are two independent events, the probability of A and B, written P(A and B) is P(A)xP(B). So if event A is the probability of a 5, P(A)=1/6 and if B is heads, P(B)=1/2 So P(A and B)=1/6 x1/2=1/12
What is -3p plus 7 equals 5-8p plus 6?
==>-3p+7=5-8p+6 ==>5p+7=11 ==>5p=4 ==>p=4/5
How do you find p(b) when p(a) is 23 p(ba) is 12 and p(a U b) is 45 and is a dependent event?
There are symbols missing from your question which I cam struggling to guess and re-insert. p(a) = 2/3 p(b ??? a) = 1/2 p(a ∪ b) = 4/5 p(b) = ? Why use the set notation of Union on the third given probability whereas the second probability has something missing but the "sets" are in the other order, and the order wouldn't matter in sets. There are two possibilities: 1) The second probability is: p(b ∩ a) = p(a ∩ b) = 1/2 → p(a) + p(b) = p(a ∪ b) + p(a ∩ b) → p(b) = p(a ∪ b) + p(a ∩ b) - p(a) = 4/5 + 1/2 - 2/3 = 24/30 + 15/30 - 20/30 = 19/30 2) The second and third probabilities are probabilities of "given that", ie: p(b|a) = 1/2 p(a|b) = 4/5 → Use Bayes theorem: p(b)p(a|b) = p(a)p(b|a) → p(b) = (p(a)p(b|a))/p(a|b) = (2/3 × 1/2) / (4/5) = 2/3 × 1/2 × 5/4 = 5/12
What is p if p plus 7 equals 13?
p + 7 = 13 p = 6
