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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)

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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)

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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.

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-1

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++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<100; ++p);

Which version you use depends upon which value you want returned from the operator, whether it is the previous value or the new value.

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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

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