Theortical Probability :
In online statistics help , In a random experiment, let there be n mutually exclusive and equally likely elementary events. Let E be an event of the experiment. If m elementary events form an event E(are favourable to E), then the probability of E, is defined as
P(E) =$\frac{m}{n}$
=$\frac{number of elements of events in E}{Total number of events in random experiment}$
Complementary events:
An event associated with a random experiment denoted Ec or not-E which happens only when E does not happen is called the complement of the event E.
P(E) + P(Ec) = 1
Probability of an impossible event :
Where there be n possible outcomes of a random experiment and no outcome favours an event, the event is called an impossible event.
P(E) = 0 then event E is called an impossible event.
Probability of a Sure Event :
When all the n possible outcomes of a random experiment favour an event, the event is called a sure event.
P(E) = 1 then the event E is a sure event.
Problems on Theortical Probability :
The following online math problems give a step by step explanation .
Q1 : Two coins of different denomination are tossed simultaneously. What is the probability of gettion (i) on head ? (ii) at least on head ? (iii) at the most one head ? (iv) no head ? (v) two heads ?
Solution : Possible outcomes are :
HH, TH, TT, i.e., 4 outcomes.
(i) P (one head ) = P(HT, TH) = $\frac{2}{4}$ = $\frac{1}{2}.
(ii) P(at least one head) = P (HT, TH, TT) = $\frac{3}{4}$.
(iii) P(at the most one head) = P(HT, TH, TT) = $\frac{3}{4}$.
(iv) P(no head) = P(TT) = $\frac{1}{4}$.
(v) P(two heads) = P(HH) = $\frac{1}{4}$.
Q2 : A fair balanced dice with six faces marked with numbers 1, 2, 3, 4, 5, 6 is thrown once. What is the prabability of getting (i) and even number ? (ii) an odd number ? (iii) a number multiple of 3 ? (iv) a prime number ? (v) a number greater than 6 ?
Solution : Possible outcomes are 1, 2, 3, 4, 5, 6, i.e., in all 6 outcomes
(i) P(an even number) = P(2, 4, 6) = $\frac{3}{6}$ = $\frac{1}{2}$
(ii) P(an odd number) = P(1, 2, 3) = $\frac{3}{6}$ = $\frac{1}{2}$
(iii) P(a number multiple of 3) = P(3, 6) = $\frac{2}{6}$ = $\frac{1}{3}$
(iv) P(a prime number) = P(2, 3, 5) = $\frac{3}{6}$ = $\frac{1}{2}$
(v)P(anumber greater than 6) = Probability of an impossible event = 0
Q3 : Cards each marked with one of the numbers 8, 9, 10, 11, 12, ......, 30 are placed in a box and mixed throughly. One card is drawn at random from the box. What is the probability of getting (i) an even number ? (ii) an odd number ? (iii) a prime number ? (iv) a number multiple of 5 ? (v) a number by 3 ?
Solution : The possible outcomes 8, 9, 10, 11, 12,............, 30 are in all 23 (total number of possible outcomes)
(i) P(an even number) = P(8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30) =$\frac{12}{23}$
(ii) P(an odd number) = P(9, 11, 13, 15,........, 29) = $\frac{11}{23}$
(iii) P(a prime number) = P(11, 13, 17, 19, 23, 29) = $\frac{6}{23}$
(iv) P(a number multiple of 5) = P(10, 15, 20, 25, 30) = $\frac{5}{23}$
(v) P(a number divisible by 3) = P(9, 12, 15, 18, 21, 24, 27, 30) = $\frac{8}{23}$
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