How many times must a man toss a fair coin so that the probability of having at least one head is more than 80%?

Answer Verified Verified
Hint: In this question , you first have to see the actual probability of having no head when the coin is tossed once i.e. ½ times. Therefore the probability of having a head can be found by 1 – p( no head). The same procedure is to be followed here.

Complete step-by-step answer:
Let the required no. of tosses be n.
P( at least one head in n toss) > 0.8 ( given in the question)
Or we can say that
P( 0 head in n toss) < 0.2……….(1)
We know that the probability of having a head or a tale in coin toss is ½ for both, then tossing a coin n times for getting 0 head would be given by \[\dfrac{1}{{{2^n}}}\].
According to equation (1), we get
\[{\left( {\dfrac{1}{2}} \right)^n} = 0.2\]
Take log on both sides
\[ \Rightarrow \log {\left( {\dfrac{1}{2}} \right)^n} = \log \left( {0.2} \right)\]
\[ \Rightarrow n\log \dfrac{1}{2} = \log \left( {.2} \right)\]
\[ \Rightarrow n( - 0.3010) \leqslant ( - 0.6989)\]
\[ \Rightarrow n \geqslant 2.3219\]
∴ approximately 3 times a coin must be tossed.

Note: In probability two events are mutually exclusive if they cannot occur at the same time. If there is occurrence of two events, one is known, then the other can be found by using the formula
P(A) = 1 – P(B), where p(A) is the unknown event.
In the above question, we used a property of log also i.e.
\[\log {\left( a \right)^n} = n\log a\]
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