Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Probability of getting a head in coin flip is 1/2. If the coin is flipped two times what is the probability of getting a head in either of those attempts?

I think both the coin flips are mutually exclusive events, so the probability would be getting head in attempt 1 or attempt 2 which is :

P(attempt1) + P(attempt2) = 1/2 + 1/2 = 1

100% probability sounds wrong? What am I doing wrong. If i apply the same logic then probability of getting at least 1 head in 3 attempt will be 1/2+1/2+1/2 = 3/2 = 1.5 which I know for sure is wrong. What do i have mixed up?

share|improve this question
    
The events are not mutually exclusive: you can get a head on the first and on the second flip –  Rookatu Mar 28 at 5:55
    
Indeed. The coin tosses are independent. That is not the same as mutually exclusive. –  Graham Kemp Mar 28 at 9:58

3 Answers 3

up vote 2 down vote accepted

You are confusing the terms "independent" and "mutually exclusive".  These are not the same thing.  In fact events cannot be both "independent" and "mutually exclusive".  It's either one, the other, or neither.

"Mutually exclusive" simply means that the two events cannot happen together. If A happens then B does not and if B happens A cannot.

"Independent" simply means that the occurrence of one event is not conditional on the occurrence of the other. The probability of A happening does not depend on whether B happens or not, and vice versa.


Let $H_n$ be the indexed event of getting a head on the $n^{th}$ flip.

Given an unbiased coin, $P(H_1)=P(H_2)=\frac 1 2$

These events are independent so $P(H_1 \cap H_2) = P(H_1)\times P(H_2)$.  The outcome of one coin toss does not influence then outcome of the other.

However they are not mutually exclusive, so $P(H_1 \cup H_2) = P(H_1)+P(H_2) - P(H_1 \cap H_2)$.  Both coins can turn up heads.

Putting it together: $$\therefore P(H_1 \cup H_2) = \frac 12 + \frac 1 2 - \frac 12 \times \frac 12 = \frac 3 4$$

share|improve this answer

Let $A$ be the event of getting a tail in both tosses, then $A'$ be the event of getting a head in either tosses. So $P(A') = 1 - P(A) = 1 - 0.5*0.5 = 0.75$

share|improve this answer

probability of having head in a coin flip is 1/2 , when you flip 2 times then the probability you have at least 1 head is equal to :

1 - P(no head) = 1 - P(2 tail) = 1 - 1/2*1/2 = 3/4 or 75%

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.