Answer :

Let the man toss the coin n times.


Let p be the probability of getting a head in a toss.


Then, q is the probability of getting a tail in a toss.


Since the coin has only two outcomes, so the probability of getting a head = 1/2



Also, p + q = 1


q = 1 – p




Let X be a random variable that represents a number of occurrence of the head in n tosses of a fair coin.


Then, the probability of getting r number of heads out of total n tosses is given by this Binomial distribution.


P (X = r) = nCrprqn-r


Substituting the value of p and q in the above equation, we get


…(i)


We need to find the number of times the man must toss a fair coin so that the probability of having at least one head is more than 80%.


We can represent it as,


P (getting atleast one head) > 80%


P (X ≥ 1) > 80%


1 – P (X < 1) > 80% [ P (X ≥ 1) = 1 – P (X < 1)]


1 – P (X = 0) > 80% [ P (X < 1) = P (X = 0)]


Put r = 0 in equation (i) and then, substituting it in the above equation.











2n > 5


Now, we need to find the minimum value of n that satisfy this inequality.


Put n = 0.


20 > 5


1 > 5


But 1 5.


Put n = 1.


21 > 5


2 > 5


But 2 5.


Put n = 2.


22 > 5


4 > 5


But 4 5.


Put n = 3.


23 > 5


8 > 5


It is true.


Thus, the minimum n that satisfy this inequality is 3.


Hence, the man should toss the coin 3 or more times.


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