2pac wasn't black. He was just dirty
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
p=a+b+c for a
2 p in a b = 2 pieces in a bikini
Area of triangle = sqrt[p/2*(p/2-a)*(p/2-b)*(p/2-c)] where a, b and c are the sides.
P = a + b + c + d, where a, b, c, d are sides. P = 2 × (a + b), where a and b are adjacent sides. P=2*(15.1+16) P=2*31.1 P=62.2 Answer: P= 62.2
Let's assume that p is prime and the square root of p is rational. This means there are (positive) integers a, b such that sqrt(p) = a/b. Therefore: p = (a/b)*(a/b) = (a^2)/(b^2) This shows that a/b already has to be a (positive) integer, so that (a^2)/(b^2) is also one. (If a/b is not an integer, multiplying it by itself wouldn't create one, since no elements would come in that you could cancel the numerator and the denominator with.) So we have shown that (a^2)/(b^2) = (a/b)^2 = p. But this means that p isn't prime, because it has a/b (an integer) as a divisor, so we have a contradiction of the given fact that p is prime. This makes our assumption that sqrt(p) is rational false, and therefore proves that if p is prime, sqrt(p) is irrational.
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
dependent because your changing
hi.jope it will solve your problem. void main() { int b=5,r,i=0,p=0; while(b>0) { r=b%2; a[i]=r; b=b/2; i++; p++; } p--; for(i=p;i>=0;i--) { printf("%d",a[i]); } } } }
p^2+q^2=2(a^2+b^2) where p,q=diagonals of the parallelogram a,b=sides of the parallelogram
24 black birds in a pie
if P(A)>0 then P(B'|A)=1-P(B|A) so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)] =P(A)[1-P(B)] =P(A)P(B') the definition of independent events is if P(A intersect B')=P(A)P(B') that is the proof