Independent Events Probability - Formula, and Solved Examples
An action is known as an event. We say that the events are independent only when the occurrence or non occurrence of any one event does not at all affect the occurrence or non occurrence probability of another event.

For Example: consider picking up a red card and then queen of heart from a given deck of cards. If we have placed back the first card in the deck then both the events are independent of each other. But in case if we have not put back the first card and then drawn the second then the probability of the second event will be depending on the probability of the first event.


In a random experiment, when two events say X and Y are independent of each other then we will have:

P ($\frac{X}{Y}$) = $P (X)$ and P ($\frac{Y}{X}$) = P (Y)

Also, when X and Y are two independent events of a random experiment, we have

P ($X \cap  Y$) = P (X) P (Y)

For proving it, we can simply use the multiplication theorem which gives us

P ( $X \cap  Y$) = P (X) P ($\frac{Y}{X}$)        …..(i)

But since X and Y are independent events so P ($\frac{Y}{X}$) = $P (Y)$        …..(ii)

From (i) and (ii), we get,

P ($X \cap  Y$) = P (X) P(Y)

Similarly we can also prove that if $X_1, X_2, …. , X_n$ are ‘n’ independent events of a random experiment then we have,

$P (X_1 \cap  X_2 \cap  … \cap X_n)$ = $P (X_1) P (X_2) … P (X_n)$

Also when the above holds true then we say that $X_1$, $X_2$, …, $X_n$ are mutually independent events and vice versa.
In words we say that the events are mutually independent if the simultaneous occurrence probability of any finite number of the events is same as the product of their respective probabilities.


Example 1: We tossed three coins. Let A be the event with all heads or all tails, B be the event of at least two tails and C be event of at most two tails. Which of the pairs of events are independent?


Possible outcomes are, $S = {TTT, TTH, THT, HTT, HTH, HHT, THH, HHH}$

According to the question, $A = {TTT, HHH}, B = {TTH, HTT, THT, TTT}$ and

Now $A \cap B = {TTT}$, $A \cap C = {HHH}$, $B \cap C = {TTH,HTT, THT}$

$P (A)$ = $\frac{2}{8}$

$P (B)$ = $\frac{4}{8}$

$P (C)$ = $\frac{7}{8}$

$P (A \cap B)$ = $\frac{1}{8}$

$P (A \cap C)$ = $\frac{1}{8}$

$P (B\cap C)$ = $\frac{3}{8}$

We can see that $P(A \cap B)$ = $P (A) P (B)$

But $P (A \cap C) \neq P (A) P(C)$ and $P (B \cap C) \neq P (B) P (C)$

So we can say that only A and B are independent events, while A & C and B & C are not independent.

Example 2: A basket contains 3 oranges, 5 apples and 2 peaches. If four fruits are drawn one after another with replacement, then find the probability that in none of the chance we got an orange.


Let $E1$, $E2$, $E3$ and $E4$ be the events of not getting an orange in first, second, third and fourth pick respectively. Since every pick is done after replacement, thus the events E1, E2, E3 and E4 are independent events.

Also, $P (E1) = P (E2) = P (E3) = P (E4)$.

Since there are 7 fruits apart from oranges, so $P (E1)$ = $\frac{7}{10}$

So the required probability is given by

P ($E1 \cap E2 \cap E3 \cap E4$) = $P (E1) P (E2) P (E3) P (E4)$ = $(\frac{7}{10})^4$