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Suppose that x is binomially distributed with size n and probability p of success what is the probability that x is greater than or equal to 1?
We have to use the expression P(X=x) = nCx px (1--p)(n--x) Here n = n and p=p and x = 1 or x>1 P(X>/=1) = 1 -- P(X</=1) So, P(X<=1) = P(X=0) + P(X=1) This gives nC0 p0 (1--p)(n-0)+ nC1 p1 (1--p)(n--1) ie (1--p)n + n p (1--p)(n--1)
Probablity of an event occurring to the probablity that it will not occor?
Let p = probability the event will occur; and q = probability the event will not occur. The relationship is p=1-q or q=1-p.
What is the value of p if p-6 equals -5?
-1
Difference between plus plus p and p plus plus?
++p increments p by 1 unit and returns the result. This is known as pre-increment.int p = 0;int q = ++p; // q=1, p=1.This is effectively the same as saying:int p = 0;p = p + 1; // p= 1.int q = p; // q = 1.p++ also increments p by 1 unit, but returns the previous value of p, not the current value. This is known as post-increment.int p = 0;int q = p++; // q=0, p=1.This is the same as saying:int p = 0;int q = p; // q = 0.p = p + 1; // p = 1.Of the two forms, ++p is marginally quicker because p++ employs a temporary variable for the return value, whereas ++p does not. As such, ++p is the preferred form for looping purposes:for( int p=0; p
If A and B are independent events then are A and B' independent?
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
HCF and LCM of p and p plus 1?
To consecutive numbers cannot have any common prime factors. Thus the HCF of p and (p+1) will be 1. To find the LCM, you multiply two numbers together and divide by the HCF. In this case, you'd do p(p+1)/1 This simplifies to p2+p So the HCF is 1 and the LCM is p2+p
What is the value of p 5(p plus 6)25?
If 5*(p + 6) = 25 then p is -1
P plus -2 equals 1?
p + (-2) = 1 so p - 2 = 1 Add two to both sides: p = 3
What is the truth table for p or q and the opposite of p and q?
P . . Q . . (P or Q)0 . . 0 . . . 00 . . 1 . . . 11 . . 0 . . . 11 . . 1 . . . 1=================P . . Q . . NOT(P and Q)0 . . 0 . . . . 10 . . 1 . . . . 11 . . 0 . . . . 11 . . 1 . . . . 0
What is x-1?
times one <P> <P>Any number multiplied by 1 will be itself. <P>1x2=2 <P>1x3=3 <P>and so on</P>
Why is the sum of the reciprocals of all of the divisors of a perfect number is equal to 2?
Suppose the perfect number P has divisors f1, f2, ... , fk, P (where all the fs are smaller than P. Then, by the definition of a perfect number, f1 + f2 + ... + fk = P So that f1 + f2 + ... + fk + P = 2P Dividing through by P, f1/P + f2/P + ... + fk/P + P/P = 2 . . . . . . . . . . . (2) Now, since f1 is a factor of P, the f1*g1 = P where g is the cofactor and so f1/P = 1/g1. That is, f1/P is the reciprocal of one of the other factors of P. Also, the last term on the left is the reciprocal of the factor 1. And therefore, f1/P + f2/P + ... + fk/P + P/P = 2 = 1/g1 + 1/g2 + ... + 1/gk + 1/1
Can you answer this 6 equals p-17 plus 1?
6 = p - 17 + 1-------------------5 = p - 17-------------------p = 22
