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**(1) An operation which can produce some well defined outcome(s) is a deterministic experiment.**

**(2) An experiment which when performed produces one of the several possible outcomes is called a random experiment.
**

Similarly, rolling an unbiased die is an example of a random experiment.

**(3) When we perform an experiment it is called a trial of the experiment.
**

If a die is thrown 5 times, then each throw is called a trial.

**(4) An outcome of a trial of an experiment is an elementary event.
**

**(5) A collection of two or more possible outcomes(elementary events) of an experiment is called a compound event.
**

**(6)An events is aid to happen in trial if any one of the elementary events(or outcomes) satisfying its conditions is an outcome.**

**(7)In n trials of a random experiments, if an event A happens m times, then the probability of happening of A is given by P(A) .
**

Head: 455, Tail: 545

Compute the probability for each event.

Number of heads = 455

Number of tells = 545

Let E be the event of getting a head:

P(E) = Number of head/ Total number of trials

P (E) = 455/ 1000

P(E) = 91/200

P(E) = 0.455

Let E

P(E

P (E

P(E

P(E

Thus , Probability of occurrence of head and tail are 0.455 and 0.545 respectively.

**(8) For any event a associated to an experiment, we have **

** For Example: **As shown above example, the value of P(E) is between 0 and 1.

**(9) If are n elementary events associated to a random experiment, then**

**
**

Two heads : 95 times

One tail : 290 times

No head : 115 times

Find the probability of occurrence of each of these events.

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** Solution: **Total no. of tosses = 500

No. of two heads appear = 95 times

No. of one heads appear = 290 times

No. of no head appear = 115 times

Let E_{1} be the event of getting two heads-

P(E_{1}) = No. of two heads appear/ No. of total tosses

P(E_{1}) = 95/500

P(E_{1}) = 19/100

P(E_{1}) = 0.19

Let E_{2} be the event of getting one tail-

P(E_{2}) = No. of one tail appear/ No. of total tosses

P(E_{2}) = 290/500

P(E_{2}) = 58/100

P(E_{2}) = 0.58

Let E_{3} be the event of getting no head-

P(E_{3}) = No. of no head appear/ No. of total tosses

P(E_{3}) = 115/500

P(E_{3}) = 23/500

P(E_{3}) = 0.23

Here .

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