Question

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is:
A) 5
B) 6
C) 7
D) 8

Answer 1

In this to find the minimum number of times a fair coin to be tossed to get at least one head. First we consider that n number of tossed is required to get at least one head and then by comparing with given condition we will find the value of n.

Complete step by step answer:
Given that a fair coin is tossed such that we get at least one head.
Probability of getting head we a fair coin is tossed = $\dfrac{1}{2}$
Now, let n be the required minimum numbers of coins tossed to get at least one head.
Probability of getting at least one head =1 – (probability of getting no head in n tossed).
Probability of getting at least one head =$1-{{\left( \dfrac{1}{2} \right)}^{n}}$
Also, given that the probability of getting at least one head is more than 99%
This implies that, Probability of getting at least one head $>\dfrac{99}{100}$
$\Rightarrow 1-{{\left( \dfrac{1}{2} \right)}^{n}}>\dfrac{99}{100}$
$\Rightarrow 1-\dfrac{99}{100}>{{\left( \dfrac{1}{2} \right)}^{n}}$
By cross multiplication, we get
$\Rightarrow \dfrac{100-99}{100}>{{\left( \dfrac{1}{2} \right)}^{n}}$
$\Rightarrow \dfrac{1}{100}>{{\left( \dfrac{1}{2} \right)}^{n}}$

$\Rightarrow \dfrac{1}{100}>\dfrac{1}{{{2}^{n}}}$
$\Rightarrow {{2}^{n}}>100$
Implies, n = 7

So, the correct answer is “Option C”.

Note: In this problem, we should not calculate the probability of getting at least one head by taking each trial to calculate the minimum number of toss required.