Question

find the probability of getting an even no. from the first 7 even natural number (hint - the probability is not 3/7 ) keep it in mind
options -1/7,6/7,7,2/7​

Answer 1

Answer:

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Step-by-step explanation:

The empty set, denoted ∅, is the set that has no element.

2. N := {1, 2, . . .}, the set of Natural numbers;

3. W := {0, 1, 2, . . .}, the set of whole numbers

4. Z := {0, 1, −1, 2, −2, . . .}, the set of Integers;

5. Q := {

p

q

: p, q ∈ Z, q 6= 0}, the set of Rational numbers;

6. R := the set of Real numbers; and

7. C := the set of Complex numbers.

option c

Answer 2

Answer:

$\huge \green\bigstar \: \rm{ANSWER \: : \: }\rm{P(E)} = 1$

Step-by-step explanation:

$\huge \red\bigstar \: \rm{TO \: FIND :}$

The probability of choosing an even number from the first 7 even natural numbers.

$\huge \orange \bigstar \: \rm{SOLUTION \: :}$

All possible outcomes are : First seven even natural number.

All possible outcomes are : 2 , 4 , 6 , 8 , 10 , 12 and 14.

Total number of possible outcomes = 7

Let E be the event of getting a even natural numbers.

Favourable outcomes are : Even number from the first seven even natural numbers.

Favourable outcomes are : 2 , 4 , 6 , 8 , 10 , 12 and 14.

Number of favourable outcomes = 7

We know that

$\rm{P(E)} = \frac{Number \: of \: outcomes \: favourable \: to \: E}{Number \: of \: all \: possible \: outcomes \: of \: the \: experiment} \\ \\ \rm{P(E)} = \frac{7}{7} \\ \\ \rm{P(E)} = \frac{ \cancel7}{ \cancel7} \\ \\ \rm{P(E)} = 1$

$\huge \blue\bigstar \: \rm{ANSWER \: : \: }\rm{P(E)} = 1$