Question

If P(E) = 0.008, what is the probability of “not E” ?​

Answer 1

Given :- If P(E) = 0.008, what is the probability of “not E” ?

Solution :-

we know that,

• P(E) + P(E') = 1 .

so, putting value of P(E) , we get,

→ P(E) + P(E') = 1

→ 0.008 + + P(E') = 1

→ + P(E') = 1 - 0.008

→ + P(E') = 1.000 - 0.008

→ + P(E') = 0.992 (Ans.)

Hence, the probability of “not E” is equal to 0.992 .

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Answer 2

SOLUTION

GIVEN

P(E) = 0.008

TO DETERMINE

The probability of “not E”

CONCEPT TO BE IMPLEMENTED

PROBABILITY

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 A, connected with the experiment be m then the probability of the event A is denoted by P (A) and defined as

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

Now the total number of possible outcomes for the event '

So the total number of possible outcomes for the event " not A " is m - n

So the probability of the event " not A " is

$\displaystyle \sf{P( \bar{A}) }$

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

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

$\displaystyle \sf{ = P( {A}) }$

Special Cases

1. For an impossible event, the number of favourable cases is zero and hence it's probability is 0

2. On the other hand for a certain event S all cases are favourable cases and hence it's probability is 1

EVALUATION

Here it is given that P(E) = 0.008

So the probability of the event “not E”

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

= 1 - P(E)

= 1 - 0.008

= 0.992

FINAL ANSWER

The probability of “not E” = 0.992

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