Probability of Independent Events

What Are Independent Events?

The exact meaning of independent events defines as the happening of one event does not affect the happening of the other. The probability of occurrence of the two events is independent. This article explains Probability of independent events along with examples.

An event E can be called an independent of another event F if the probability of occurrence of one event is not affected by the occurrence of the other.

Suppose two cards are drawn one after the other. The outcome of the draws is independent if the first card is put back into the pack of cards before the second draw. If the cards are not replaced back then the events are not independent.

Consider one more example. Suppose a bag contains 5 white and 3 red marbles. Two marbles are drawn from the bag one after another. Consider the events A = Drawing a white marble in the first draw. B = Drawing a red marble in the second draw. If the marble drawn in the first draw is replaced back in the bag, then A and B are independent events because P(B) remains the same whether we get a white marble or a red marble in the first draw. 

If A and B are two independent events associated with a random experiment, then 

P(A/B) = P(A)  and P(B/A) = P(B) and vice versa.

If S is the sample space of the random experiment, A and B are any two events defined in this sample space. The two events A and B are said to be independent, that is

If P (A / B) = P (A / B’) = P (A) or

P (B / A) = P (B / A’) = P (B) and

P (AB) = P (A) * P (B)

Theorem 1 : If A and B are two independent events associated with a random experiment, then P(A⋂B)  = P(A) P(B)

Probability of simultaneous occurrence of two independent events is equal to the product of their probabilities. 

Theorem 2: If A1,A2,…An are independent events associated with a random experiment, then P(A1⋂A2⋂A3….⋂An) =  P(A1) P(A2)P(A3)….P(An)

How are independent events and mutually exclusive events different?

Foundation for comparison Independent events Mutually exclusive events
Definition The events are independent if the occurrence of one doesn’t result in any change in the occurrence of another event. The events are mutually exclusive if they don’t occur simultaneously.
Impact The occurrence of one event doesn’t impact the occurrence of the other. The occurrence of one event results in the non-occurrence of the other event.
Formula P ( AB) = 0 P (AB) = P (A) * P (B)
Venn diagram [Sets] No overlapping Overlapping is present.

Also Read

Dependent Events in Probability

Mutually Exclusive events in Probability

Probability Problems

Probability Of Independent Events Examples

Example 1: Let E and F be two independent events. The probability that exactly one of them occurs is 11 / 25 and the probability of none of them occurring is 2 / 25. What is the probability of occurrence of the event T?

Solution:

P (E ∩ F) − P (E ∩ F) = 11 / 25 …(i)

[i.e. only E or only F]

Neither of them occurs = 2 / 25

⇒P(E\overline EF\overline F) = 2 / 25 ….(ii)

From Eq. (i), P (E) + P (F ) – 2 P (E∩F) = 11 / 25 ….(iii)

From Eq. (ii), (1 − P(E)) (1 − P(F)) = 2 / 25

⇒ 1 − P (E) − P (F) + P (E) * P (F) = 2 / 25 ….(iv)

From Eqs. (iii) and (iv),

P (E) + P (F) = 7 / 5 and P (E)* P (F) = 12 / 25

∴ P (E) * [7 / 5 − P (E)] = 12 / 25

⇒ (P (E))2 − [7 / 5] P (E) + 12 / 25 = 0

⇒ [P (E) − 3 / 5] [P (E) − 4 / 5] = 0

∴ P (E) = 3 / 5 or 4 / 5

⇒ P (F) = 4 / 5 or 3 / 5

Example 2: Let E and F be two independent events. If the probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2. Then, find P (E) and P (F).

Solution:

Both E and F happen ⇒ P (E ∩ F) = 1 / 12 and neither E nor F happens →P(E\overline EF\overline F) = 1 / 2

But for independent events, we have P (E ∩ F) = P (E) P (F) = 1 / 12 …(i) and

P(E\overline EF\overline F) = P(E)P(F)P(\overline E) * P (\overline F)

= { 1 − P (E)} { 1 − P (F)}

= 1 − P (E) − P (F) + P (E) P (F)

⇒P (E) + P (F) = 1 − 1 / 2 + 1 / 12 = 7 / 12 …(ii)

On solving Eqs. (i) and (ii), we get

either P (E) = 1 / 3 and P (F) = 1 / 4 or P (E) = 1 / 4 and P (F) = 1 / 3

Example 3: Let A and B be two events such that P(AB)=16,P(AB)=14P\overline{(A\cup B)}=\frac{1}{6},P(A\cap B)=\frac{1}{4} and P(Aˉ)=14,P(\bar{A})=\frac{1}{4}, where Aˉ\bar{A} stands for complement of event A. Then events A and B are       

A) Independent but not equally likely

B) Mutually exclusive and independent

C) Equally likely and mutually exclusive

D) Equally likely but not independent

Solution: 

P(AB)=16;P(AB)=14,P(Aˉ)=14P(A)=34,P(AB)=1P(AB)=1P(A)P(B)+P(AB)16=14P(B)+14P(B)=13P\left( \overline{A\cup B} \right)=\frac{1}{6};\,\,P\left( A\cap B \right)=\frac{1}{4}, \\ P\left( {\bar{A}} \right)=\frac{1}{4}\Rightarrow P\left( A \right)=\frac{3}{4}, \\ P\left( \overline{A\cup B} \right)=1-P\left( A\cup B \right)=1-P\left( A \right)-P\left( B \right)+P\left( A\cap B \right) \\ \frac{1}{6}=\frac{1}{4}-P\left( B \right)+\frac{1}{4}\\ P\left( B \right)=\frac{1}{3}

Since 

P(AB)=P(A)P(B) and P(A)P(B)P\left( A\cap B \right)=P\left( A \right)P\left( B \right) \text \ and \ P\left( A \right)\ne P\left( B \right)

A and B are independent but not equally likely.

Example 4: The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2. Then what is P(A\overline A) + P(B\overline B)?

Solution:

Given :

P (A ∪ B) = 0.6, P (A ∩ B) = 0.2

∴ P (A ∪ B) = P(A) + P (B) − P (A ∩ B)

0.6 = P (A) + P (B) − 0.2

⇒ 0.6 + 0.2 = P (A) + P (B)

⇒ P (A) + P (B) = 0.8

Step 2:

P(A\overline A) + P(B\overline B) = (1 − P(A)) + 1 − P(B))

⇒ 2 − (P (A) + P (B))

⇒ 2 − 0.8 = 1.2

Example 5: If E and F are independent events such that 0 < P (E) < 1 and 0 < P (F) < 1, then write the conditions for them to be independent.

Solution:

P (E ∩ F) = P (E) * P (F)

Now, P (E ∩ Fc) = P (E) − P (E ∩ F) = P (E) [1 − P (F)]= P (E) * P(Fc) and

P (Ec ∩ Fc) = 1 − P (E ∪ F) = 1 − [P (E) + P (F) − P(E ∩ F)

= [ 1 − P (E)] [1 − P (F)] = P (Ec) P (Fc)

Also P (E / F) = P (E) and P (Ec/Fc) = P (Ec)

⇒ P (E / F) + P (Ec / Fc) = 1

Example 6: The odds against a certain event is 5 : 2 and the odds in favour of another event is  6 : 5. If both the events are independent, then the probability that at least one of the events will happen is  

Solution: 

Let A and B be two given events. 

The odds against A are 5:2, therefore P (A) = 2 / 7 .                    

The odds in favour of B  are 6:5, therefore P (B) = 6 / 11.

The required probability is 

=1P(Aˉ)P(Bˉ)=1(127)(1611)=5277.=1-P(\bar{A})\,P(\bar{B}) \\ =1-\left( 1-\frac{2}{7} \right)\,\left( 1-\frac{6}{11} \right)\\=\frac{52}{77}.

Example 7: If two events A and B are such that P(A+B)=56,P(AB)=13P(Aˉ)=12,P\,(A+B)=\frac{5}{6}, P\,(AB)=\frac{1}{3}\, P\,(\bar{A})=\frac{1}{2},

then the events A and B are

A) Independent

B) Mutually exclusive

C) Mutually exclusive and independent

D) None of these

Solution:

P(A+B)=P(A)+P(B)P(AB)56=12+P(B)13P(B)=46=23P(A).P(B)=12×23=13=P(AB)P(A+B)=P(A)+P(B)-P(AB) \\ \Rightarrow \frac{5}{6}=\frac{1}{2}+P(B)-\frac{1}{3}\\ \Rightarrow P(B)=\frac{4}{6}=\frac{2}{3}\\ P(A)\,.\,P(B)=\frac{1}{2}\times \frac{2}{3}=\frac{1}{3}=P(AB)

Hence events A and B  are independent.

 Example 8: Let A and B are two independent events. The probability that both A and B occur together is 1 / 6 and the probability that neither of them occurs is 1 / 3. The probability of occurrence of A is 

Solution: 

P(AB)=16 and P(AcBc)=13P(AB)c=P(AcBc)=131P(AB)=13P(AB)=23P(A\cap B)=\frac{1}{6} \text \ and \ P({{A}^{c}}\cap {{B}^{c}})=\frac{1}{3} P{{(A\cup B)}^{c}}=P({{A}^{c}}\cap {{B}^{c}})=\frac{1}{3} \\ 1-P(A\cup B)=\frac{1}{3} \Rightarrow \,P(A\cup B)=\frac{2}{3}

But 

P(AB)=P(A)+P(B)P(AB)P(A)+P(B)=56(i) because A and B are independent eventsP(AB)=P(A)P(B)P(A)P(B)=16[P(A)P(B)]2=[P(A)+P(B)]24P(A)P(B)=253646=136P(A)P(B)=±16(ii) Solving (i) and (ii), we get P(A)=12 or 13.P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ \Rightarrow P(A)\,+P(B)=\frac{5}{6} \rightarrow (i) \text \ because \ A \ and \ B \ are \ independent \ events \\ P(A\cap B)\,=\,P(A)\,P(B) \\ P(A)\,P(B)=\frac{1}{6} \\ {{[P(A)\,-\,P(B)\,]}^{2}}={{[P(A)+P(B)]}^{2}}-4P(A)\,P(B) \\=\frac{25}{36}-\frac{4}{6}=\frac{1}{36} \\ \Rightarrow P(A)-P(B)=\pm \,\frac{1}{6} \rightarrow (ii) \\ \text \ Solving \ (i) \ and \ (ii), \ we \ get \ P(A)=\frac{1}{2} \text \ or \ \frac{1}{3}.