Self Online Study  Mathematics  ProbabilityThe concepts of probability and various result on it were discussed in the class XI. In this chapter, we shall be dealing with problems based on multiplication theorem on probability for independent events and those on conditional probability. Independent Events : Two events ${\mathrm{E1}}_{\mathrm{}},{E}_{2}$are said to be independent if the occurrence of one does not depend upon the occurrence of the other. Example Suppose we tossed two unbiased coins. Let ${E}_{1}$= event of getting a head on the first coin. and ${E}_{}$_{ 2} = event of getting a head on the second coin. Then it is clear that the occurrence of a head on the second coin does not dependent upon the occurrence of a head on the first coin. Hence ${\mathrm{E1}}_{},{E}_{}$_{ 2} are independent events. Multiplication Rule For Independent Events $\begin{array}{l}{\text{If E}}_{}\end{array}$_{ 1} and E_{ 2} are independent events then P( E_{ 1} and E_{ }2 )=P( E_{1} ∩ E_{ 2} )=P( E_{ 1} )×P( E_{ 2} ) Consider tossing a coin three times in a row. Since each of the throws is independent of the other two, we consider all 8 (= 23) possible outcomes as equiprobable and assign each the probability of 1/8. Here is the sample space of a sequence of three tosses: $$\text{{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}}\text{.}$$ There are $${\text{2}}^{\text{8}}$$possible events, but we are presently interested in, say, two: $$\text{A = {HHH, HTH, THH, TTH} andB = {HHH, HHT, THH, THT}}\text{.}$$ A is the sequence of tosses in which the third one came up heads. B is the event in which heads came up on the second toss. Since each contains 4 outcomes out of the equiprobable 8, $$\text{P(A) = P(B) = 4/8 = 1/2}\text{.}$$ The result might have been expected: 1/2 is the probability of the heads on a single toss. Are events A and B independent according to the definition? Indeed they are. To see that, observe that $$\text{A}\cap \text{B = {HHH, THH},}$$ the event of having heads on the second and third tosses $$\text{.}$$P(A ∩B) = 2/8 = 1/4 . Further, letâ€™s find the conditional probability P(AB): $$\begin{array}{l}\text{P(AB)= P(A}\cap \text{B) / P(B)}\\ \text{= 1/4 / 1/2}\\ \text{= 1/2}\\ \text{= P(A)}\text{.}\\ \end{array}$$ So that P(AB) = P(A) and, according to the definition, events A and B are independent, as expected. This is in fact always the case. Assume we run a sequence of (independent) experiments with, among others, two possible outcomes x and y with probabilities P(x) = p and P(y) = q. The event that in a sequence of experiments the first outcome happened to be x has the probability of p because something happens on every trial with the probability of 1 and the combinatorial product rule applies. Similarly, the event of having the outcome of y on the second trial (in any sequence of experiments) has the probability of q. Using random variables V_{1} and V_{2} for the outcomes of the first and second experiments, we may express this in the following manner: $$\begin{array}{l}{\text{P(V}}_{}\end{array}$$_{ 1} = x) = p andP(V_{ 2} = y) = q. If A and B are the corresponding events, P(A) = p, P(B) = q. The event A n B of having x on the first experiment and y on the second has the probability of 1 / pq. It follows that $$\begin{array}{l}\text{P(AB)= P(A}\cap \text{B) / P(B)}\\ \text{= 1/pq / 1/q}\\ \text{= 1/p}\\ \text{= P(A)}\\ \\ \end{array}$$ making the events A and B independent.Diagrams that picture the sample space, S, as an area in the plane and events as areas within S are called Venn Diagrams. In the first Venn diagram (left), events A and B are mutually exclusive. In the second Venn diagram (right), events A and B are not mutually exclusive.

