Question

What is the value of p(E)+p(notE).

Answer 1

P(E) + P(not E) = 1

Given :

P(E) + P(not E)

To find :

The value of P(E) + P(not E)

Solution :

Step 1 of 2 :

Define probability of an event

In any random experiment if the total number of elementary ( simple) events in the sample space be n ( a finite number) among which the number of elementary events favourable to an event E , connected with the experiment be m then the probability of the event E is denoted by P (E) and defined as

$\displaystyle \sf{}P(E) = \frac{m}{n}$

Now the total number of possible outcomes = n and the total number of possible outcomes for the event E is m

Step 2 of 2 :

Find the value of P(E) + P(not E)

Ē (not E) is the event of not occurring E

Then total number of possible outcomes for the event Ē is n - m

Thus we get

P(not E)

$\displaystyle \sf = P(\bar{E})$

$\displaystyle \sf = \frac{n - m}{n}$

$\displaystyle \sf = 1 - \frac{m}{n}$

$\displaystyle \sf = 1- P({E})$

Hence we get

P(E) + P(not E)

$\displaystyle \sf = P(E) + P(\bar{E})$

$\displaystyle \sf = P(E) + 1 - P(E)$

= 1

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