It can be carried out any number of times under the identical condition.
In these experiments the possible set of outcomes is known.
But the outcome of any individual trial is unknown till it is executed.
Tossing a coin: $\{T, H\}$
Throwing a die: $\{1,2,3,4,5,6\}$
Tossing a coin three times and noting the sequence of outcomes
$\{H H H, H H T, H T H, H T T, T H H, T H T, TTH,TTT\}$
Notation: $\Omega$ or $S$
and this is called the sample space
$\therefore$ Number of sample points
$2+3+5+3+11+16=30$
The random experiment is the following
Suppose in a bus stop there are four passengers. Each passenger may have 1,2 or no bags. You come to the bus stop and pick up 1 or 2 or 3 or 4 passengers, and each passenger takes his/her bags
Describe the sample space as a pair of numbers (# of passengers, # of bags)
$\Omega =$ $\{(1,0),(1,1),(1,2),(2,0),(2,1),(2,3),(2,4),$$(3,0),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$ $(4,0),(4,1),(4,2),(4,3),$ $(4,4),(4,5),(4,6),(4,7),(4,8)\}$
$|\Omega|=3+5+7+9=24$
Example 1
$n$ can be any positive integer.
$\therefore \Omega=\{1,2,3, \cdots\} \quad $ infinite
Example
Note
$(A, B C), (B, A C), (C, A B)$
$(A B, C), (A C, B), (B C, A)$
$\therefore 6 \text{ possible ways}$
$\therefore$ Four possibilities
$\therefore\text{Number of possibilities} = 1$
$\begin{array}{r}(1,9)(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2)(9,1) \end{array}$ -
$9$ Possibilities
Now consider 3 boxes & none will remain empty
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$\therefore$ Total no of possibilities
$ 1+2+\cdots+8=\frac{8 \times 9}{2}$
$ =36 \text { different ways} $
Number of solution is $ ^{(10-1)} C_{(3-1)}= ^9 C_2$
$=\frac{9 !}{2 ! 7 !}=\frac{8 \times 9}{2}=36$
Therefore in general if there are $n$ identical ball to be put in $k$ boxes such that none of the box is empty, then the number of possible ways $={ }^{n-1} C_{k-1}$
Answer
possible way are $ ^{14} C_2$
Consider the following, We have 10 identical balls to be put in 3 boxes such that any box can remain empty. what are the number of possibilities?
$n+k$ elements can be put in $k$ boxes such that none is empty is $ ^ {n+k-1} C_{k-1}$
$(0,3) \quad (1,2) \quad (2,1)\quad(3,0)$
$(1, 4 )$
$(1-1, 4-1)\rightarrow (0,3)$
$(2,3)$
$(2-1, 3-1)\rightarrow (1,2)$