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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
How do you solve 6 21 b 6 21 b?
how to solve 7 + p = p +7
B is 7 p is 6 b is 5?
Think Snooker: Black is 7, Pink is 6, Blue is 5. And there's ya answer. From Nightfire-Player. (no account here yet)
What is the probability of rolling either a 5 or 6 or 8 or 9 before rolling a 7 with two dice?
Consider the three events: A = rolling 5, 6, 8 or 9. B = rolling 7 C = rolling any other number. Let P be the probability of these events in one roll of a pair of dice. Then P(A) = P(5) + P(6) + P(8) + P(9) = 18/36 = 1/2 P(B) = P(7) = 6/36 = 1/6 and P(C) = 1 - [P(A) + P(B)] = 1/3 Now P(A before B) = P(A or C followed by A before B) = P(A) + P(C)*P(A before B) = 1/2 + 1/3*P(A before B) That is, P(A before B) = 1/2 + 1/3*P(A before B) or 2/3*P(A before B) = 1/2 so that P(A before B) = 1/2*3/2 = 3/4
What is P(A and B)(A) 2/7(B) 4/7(C) 5/7(D) 3/7?
What is the correct answer?
If P(A)=1/5 and P(B)=3/7 and P(A∩B)=2/17 are A and B independent or dependent?
dependent because your changing
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 PA and B for PA equals 30 and PB equals 55?
P = 5 A = 6 B = 11 Or.... P = 1 A = 30 B = 55
What is -3p plus 7 equals 5-8p plus 6?
==>-3p+7=5-8p+6 ==>5p+7=11 ==>5p=4 ==>p=4/5
What is an algebraic expression for the following word phrase 6 times the difference of b and p?
6*abs(b - p) which can also be written as 6*|b - p|
What are prime triplet numbers?
A prime triplet contains a pair of twin primes (p and p + 2, or p + 4 and p+ 6), a pair of cousin primes (pand p + 4, or p + 2 and p + 6), and a pair of sexy primes (p and p + 6).
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
