Question

if E be an event then P(E)+P(notE)=.......(and how)​

Answer 1

Answer:

If E be an event then P(E)+P($\bar E$) = 1

Step-by-step explanation:

If E is an event, then P(E) is the probability of that event to happen, and $P(\bar{E})$ is the probability of not happening that event.

Let S be the sample space, then n(S) bethe total number of outcomes.

n(E) be the number of favourable outcomes.

Then,

$P(E) = \frac{n(E)}{n(S)}$....(1)

Similarly,

$n(\bar{E})$ be the number of unfavourable outcomes out of sample space 'S'.

Then,

$P(\bar{E}) = \frac{n(\bar{E})}{n(S)}$....(2)

But as number of unfavourable outcomes is nothing but the difference between the total number of outcomes to the number of favourable outcomes.

i.e. $n(\bar{E}) = n(S) - n(E)$...(3)

Substituting equation (3) in (2)

$\implies P(\bar{E}) = \frac{n(S) - n(E)}{n(S)}$...(4)
Now add equations (4) and (1) we get:

$P(E) + P(\bar{E}) = \frac{n(E)}{n(S)} +\frac{n(S)-n(E)}{n(S)}$

$\implies P(E) + P(\bar{E}) = \frac{n(E)}{n(S)} +\frac{n(S)}{n(S)} -\frac{n(E)}{n(S)} \\\implies P(E) + P(\bar E) = \frac{n(E)}{n(S)} - \frac{n(E)}{n(S)} +1$

⇒ P(E) + P($\bar E$) = 1

Therefore, if E be an event then P(E)+P($\bar E$) = 1

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Answer 2

Answer:

The value of P(E) + P(not E) is 1.

Step-by-step explanation:

Probability of an event:-

• The probability of an event refers to the event that will surely occur.
• Mathematically, the probability of an event that will occur lies between 0 and 1.
• This means the probability of any event cannot be less than 0 and greater than 1.
• Symbolically, the probability of event E is represented by P(E).

Step 1 of 1

Let us consider the probability of an event (E) happening or occurring be p, i.e.,

P(E) = p

Then, the probability of an event (E) that does not occur will be 1 - p, i.e.,

P(not E) = 1 - p

Now,

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

                       = p + 1 - p

                       = 1

Thus, the value of P(E) + P(not E) is 1.

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